Mathematics for Grade XII 1
Modul
Mathematics for Compulsory Program
Title page
For Level XII
SMA ABBS Surakarta
Academic Year 2021/2022
Tri Wijayanti, M.Pd.
Farida Leni Kusumawati, S.Pd.
Mathematics for Grade XII 2
S M A A B B S S U R A K A R T A , I N D O N E S I A
© 2 0 2 1
Mathematics for Grade XII 3
P R E F A C E
Alhamdulillah, all praises is due to
Allah swt. Shalawat and salam may
always be given to our phrophet
Muhammad pbuh.
To all students, this textbook is
arranged to facilitate you study
mathematics. Learning is an
interesting thing. Our insight and
understanding grow and grow. As the
tree continues to be bigger, our
concept and understanding about
science will develop too.
Every time we learn, starting with the
intention that we want to get the
benefits of science, and next we can
help others with the science . That's
when your tree is becoming fruitful.
We need patience and perseverance
to study everything.
We hope you will be successful in the
world and in the hereafter. Have a
good study.
Teacher
Mathematics for Grade XII 4
CONTENTS
Title Page …………………………………………………………………………
Preface ...………………………………………………………………………….
Contents …………………………………………………………………………..
Core and Standard Competence ………………………………………………….
1
3
4
6
Chapter1 Three Dimensional Space – 8
Projection ....................................................................................... 9
Finding a Distance in Solid Shapes………………………………. 12
Chapter Review Exercises ............................................................. 38
Angles in Space …………………………………………………. 42
Chapter 2 Statistics – 49
Measuring the Center of Data ....................................................... 50
Measures of Position Data Cluster ................................................ 56
Measures of Spread ....................................................................... 60
Chapter Review Exercises ............................................................. 67
Chapter 3 Probability I – 71
Filling Possibility Rules ................................................................. 72
Factorial Notation .......................................................................... 78
Permutation .................................................................................... 81
Combination ................................................................................... 85
Binomial Newton ........................................................................... 87
Chapter Review Exercises ............................................................. 89
Mathematics for Grade XII 5
Chapter 4 Probability II – 91
Probability of an Event .................................................................. 92
Probability of the Combination of Two Mutually Exclusive Events
The Probability of Two Independent Events .................................
94
96
The Probability of Conditional Events .......................................... 99
Bibliography
Chapter Review Exercises .............................................................
...........................................................................................
106
108
Mathematics for Grade XII 6
Core and Standard Competence
Kompetensi Inti :
KI 1 : Menghayati dan mengamalkan ajaran agama yang dianutnya
KI 2 : Menghayati dan mengamalkan perilaku jujur, disiplin, tanggungjawab,
peduli (gotong royong, kerjasama, toleran, damai), santun, responsif dan
pro-aktif dan menunjukkan sikap sebagai bagian dari solusi atas berbagai
permasalahan dalam berinteraksi secara efektif dengan lingkungan sosial
dan alam serta dalam menempatkan diri sebagai cerminan bangsa dalam
pergaulan dunia
KI 3 : Memahami, menerapkan, dan menganalisis pengetahuan faktual,
konseptual, prosedural, dan metakognitif berdasarkan rasa ingin tahunya
tentang ilmu pengetahuan, teknologi, seni, budaya, dan humaniora
dengan wawasan kemanusiaan, kebangsaan, kenegaraan, dan peradaban
terkait penyebab fenomena dan kejadian, serta menerapkan pengetahuan
prosedural pada bidang kajian yang spesifik sesuai dengan bakat dan
minatnya untuk memecahkan masalah
KI 4 : Mengolah, menalar, dan menyaji dalam ranah konkret dan ranah abstrak
terkait dengan pengembangan dari yang dipelajarinya di sekolah secara
mandiri, bertindak secara efektif dan kreatif, serta mampu menggunakan
metoda sesuai kaidah keilmuan
Kompetensi Dasar
3.1 Mendeskripsikan jarak dalam ruang (antar titik, titik ke garis, dan titik ke bidang)
4.1 Menentukan jarak dalam ruang (antar titik, titik ke garis, dan titik ke bidang)
3.2 Menentukan dan menganalisis ukuran pemusatan dan penyebaran data yang disajikan
dalam bentuk tabel distribusi frekuensi dan histogram
4.2 Menyelesaikan masalah yang berkaitan dengan penyajian data hasil pengukuran dan
pencacahan dalam tabel distribusi frekuensi dan histogram
3.3 Menganalisis aturan pencacahan (aturan penjumlahan, aturan perkalian, permutasi, dan
kombinasi) melalui masalah kontekstual
Mathematics for Grade XII 7
4.3 Menyelesaikan masalah kontekstual yang berkaitan dengan kaidah pencacahan (aturan
penjumlahan, aturan perkalian, permutasi, dan kombinasi)
3.4 Mendeskripsikan dan menentukan peluang kejadian majemuk (peluang kejadiankejadian saling bebas, saling lepas, dan kejadian bersyarat) dari suatu percobaan acak
4.4 Menyelesaikan masalah yang berkaitan dengan peluang kejadian majemuk (peluang,
kejadian-kejadian saling bebas, saling lepas, dan kejadian bersyarat)
Mathematics for Grade XII 8
Chapter 1 THREE DIMENSIONAL SPACE
Figure 1.1
In this chapter, we will
discuss three-dimensional
space. The topics covered
here include how draw a
number of threedimensional objects, or
solid shapes. Drawing
solid shapes requires
ample imagination and
visualization.
This chapter will give you the basic skills for drawing simple solid shaper such as
cube, boxes (cuboids), and pyramids. Our drawing lesson here will be different
from that in Art, since it is intended to understand mathematical geometry. This
drawing skill will be particularly useful for those who aspire to pursue fields which
ground their theoriesand subject-matters in three-dimensional space, such as
architecture (Figure 1.1), liberal arts, and civil engineering. For those whi aren’t too
interested on such fields,you can consider our discussion in this chapter as a
stimulating exercise for your imagination and visualization skills.
Sesungguhnya rumah yang mula-mula dibangun untuk (tempat beribadat) manusia,
ialah Baitullah yang di Bakkah (Mekah) yang diberkahi dan menjadi petunjuk bagi
semua manusia. (Q.S. Ali ‘Imran ayat 49)
Chapter 1
What is the purpose of lesson?
3.1 Mendeskripsikan jarak dalam ruang (antar titik, titik ke garis,
dan titik ke bidang)
4.1 Menentukan jarak dalam ruang (antar titik, titik ke garis, dan
titik ke bidang)
Mathematics for Grade XII 9
A. Projection
1. Projection of a Line on a Line
The projection line segment ܱܲ on line ݃ is segment ܱܲᇱ
(Figure 1.1)
Figure 1.1
The projection of point ܲ on line ݃ is the point ܲ′ so that ܲܲ′ ⊥ ݃ .
(See Figure 1.2) The projection of line segment AB on line ݃ is line
.′ܤ′ܣ segment
Figure 1.2
If the line segment AB with a length of an angle of ߠ with line ݃, then the
length of projection of AB (that is ܤܣ (′is ܽ cos ߠ) .Figure 1.3)
Figure 1.3
2. Projection of a Point on a Line
Projection of a point ܲ on a plane ߙ is the pass-through point of a
perpendicular straight line originating from ܲ on plane ߙ) .Figure 1.4)
ܲ′ = Projection of ܲ on ߙ
ܲܲ′ = Projector or the distance between point ܲ and plane ߙ
ߙ = Plane of projection
ߙ ⊥ ′ܲܲ
Mathematics for Grade XII 10
Figure 1.6 Figure 1.7
Figure 1.4
3. Projection of a Line on a Plane
Consider Figure 1.5 . If all the points on line segment ܤܣ are projected
onto plane ߙ ,then all the projectors will be located on a single plane (the
projector plane) and all the projectors will be located on a single line
segment ܣ′ܤ .′This means projection of a line segment ܤܣ on plane ߙ is a
.′ܤ′ܣ segment line
Figure 1.5
If line segment ܤܣ is perpendicular to plane ߙ ,then its projection on plane ߙ
is just a single point, namely point ܤ which is located on the projection plane
(plane ߙ) (Figure 1.6). if line segment ܤܣ passes through plane ߙ at ܤ ,then the
projection of point ܣ on plane ߙ is point ܣ ′and the projection of point ܤ on
plane ߙ is point ܤ itself. (Figure 1.7). hence, the projection of line segment ܤܣ
on plane ߙ is ܣ
.ܤᇱ
Projection of any straight line a planar surface is generally also a
straight line
Mathematics for Grade XII 11
Example 1
Consider a cube ܦܥܤܣ .ܪܩܨܧ having an edgelength of 5 cm. (Figure 1.8)
a. Find the projection and projection length of:
(i) line ܧܣ on plane ܨܩܥܤ
(ii) line ܧܣ on plane ܦܥܤܣ
(iii) line ܩܣ on plane ܦܥܤܣ
b. Find the projection and projection area of
plane ܧܩܥܣ on plane ܦܥܤܣ Figure 1.8
Solution
a. (i) Projection of ܧܣ on plane ܨܩܥܤ is ܨܤ ,length of ܨܤ = 5 cm.
(ii) Projection of ܧܣ on plane ܦܥܤܣ is point ܣ ,length of ܣܣ = 0 cm.
(iii) Projection of ܩܣ on plane ܦܥܤܣ is ܥܣ ,length of ܥܣ = 5√2 cm.
b. Projection of plane ܧܩܥܣ on plane ܦܥܤܣ is line ܥܣ .
Area of projection (line ܥܣ = (0 cm2
.
1. Given a box ܦܥܤܣ .ܪܩܨܧ
a. Find the projection of ܧܤ
and ܪܥ on plane ܦܥܤܣ .
b. Find the projection of ܧܤ on
ܨܪܦܤ plane
2. On a cube ܦܥܤܣ .ܪܩܨܧ ,find
the projection of:
ܦܥܤܣ on ܦܨ .a
ܨܩܥܤ on ܦܨ .b
c. EC on ܩܪܦܥ
ܪܩܨܧ on ܥܣ .d
ܦܥܤܣ on ܣܧ .e
ܧܩܥܣ on ܨܧ .f
3. Given a cube ܦܥܤܣ .ܪܩܨܧ and
a point ܲ that is located in the
middle of ܩܥ .Find the
projection of:
a. ܣ ܲon the base plane
ܧܪܦܣ onܲ ܣ .b
ܨܪܦܤ onܲ ܤ .c
ܪܦܤ onܲ ܥ .d
ܨܪܦܤ on ܨܣ .e
4. Given a cube ܦܥܤܣ .ܪܩܨܧ ,
ܣ ܲ6 cm, and point ܲ is the
midpoint of ܪܩ .Draw:
a. the projection of ܣ ܲon
the base plane
b. the projection of ܣ ܲon
ܧܪܦܣ
c. the projection of ܤ ܲon
ܨܪܦܤ
d. the projection of ܥ ܲon
ܨܪܦܤ
Classroom Activities 1
Mathematics for Grade XII 12
B. Finding a Distance in Solid Shapes
1. The Distance between Two Points
The distance is the shortest line that can be drawn between the two points.
Hence, the distance between point A and point B is the length of line
segment AB (Figure 1.9)
Figure 1.9
Example 2
A cube ܦܥܤܣ .ܪܩܨܧ has an edge-length of 6 cm. Given that ܲ is the
intersection point of the top face’s diagonal lines, calculate the distance
between point ܲ and ܣ) .Figure 1.10)
Figure 1.10
Solution
Observe that ∆ܧܣ ܲis a right triangle at ܧ) ܧܣ ⊥ plane ܪܩܨܧ ,(
ଶܲܧ + ଶܧܣ = ଶܲܣ
= 6
ଶ + ቀ
ଵ
ଶ
ቁܩܧ
ଶ
= 36 + ൫3√2൯
ଶ
= 54
ܣ = ܲ3√6 cm
2. The Distance between a Point and a Line
If a point ܲ and a line ݃ are both located on a plane ߙ ,then the distance
between the point ܲ and line ݃ can be determined using the following
steps.
a. Draw a line ℎ that passes through ܲ and is perpendicular to ݃
Mathematics for Grade XII 13
b. Suppose that the lines ݃ and ℎ intersect at ܴ. This point ܴ then is the
projection of ܲ on ݃. ܴܲ is the distance between line ݃ and point ܲ.
(Figure 1.11)
Figure 1.11
If line ݃ is located on plane ߙ while point ܲ is located outside of ߙ ,then
the distance between point ܲ and line ݃ canbe determined using the
following steps.
a. Draw a line ܲܳ at a right angle with plane ߙ .
b. Draw a line ܴܳ at a right angle with line ݃.
c. ܴܲ is the distance between point ܲ and line ݃. (Figure 1.12)
Figure 1.12
Example 3
A box ܦܥܤܣ .ܪܩܨܧ has a lenght of 8 cm, a width of 6 cm, and a height
of 6 cm. Suppose that point ܲ is the intersection point of the top plane’s
diagonals ܪܨ and ܩܧ ,point ܴ is the midpoint of line segment ܪܧ ,and
point ܳ is the midpoint of line segment ܦܣ) .Figure 1.13)
a. Find the distance between point ܲ and line ܦܣ
b. Find the distance between point ܥ and line ܪܧ .
Figure 1.13
Mathematics for Grade XII 14
Solution
a. Point ܲ is outside of plane ܧܪܦܣ ,hence the distance between ܲ and
line ܦܣ can be determined by the following steps:
Draw a line ܴܲ ⊥ ܪܧ
Draw a line ܴܳ ⊥ ܦܣ
ܲܳ is the distance between point ܲ and line ܦܣ
ܲܳଶ = ܴܲଶ + ܴܳଶ
= 4ଶ + 6ଶ
ܴܲ = √52
= 2√13 cm
b. Line ܪܧ is on plane ܧܪܦܣ ,ܦܥ ⊥ ܧܪܦܣ and ܪܦ ⊥ ܪܧ .Therefore,
the distance between point ܥ and line ܪܧ is ܪܥ .
ଶܪܦ + ଶܦܥ = ଶܪܥ
= 8ଶ + 6ଶ
= 100
ܪܥ = 10 cm
3. The Distance between a Point and a Plane
The distance between a point ܲ and plane ߙ ,if ܲ is located on plane ߙ ,
is 0. If point ܲ is located outside of plane ߙ ,then the distance between ܲ
and ߙ can be determined as follows:
Draw a line ݃ that passes through point ܲ and is at a right angle
(perpendicular) to plane ߙ .Suppose that ݃ passes through ߙ at ܳ. ܲܳ is
the distance between point ܲ and plane ߙ) .Figure 1.14)
Figure 1.14
Example 4
Given a cube ܦܥܤܣ .ܪܩܨܧ with an edge length of 6 cm. Determine the
distance between point ܤ and plane ܥܨܣ .
Solution
Look at the cube ܦܥܤܣ .ܪܩܨܧ) Figure 1.15).
Mathematics for Grade XII 15
Figure 1.15
Point ܤ is located on plane ܦܥܤܣ .ܪܩܨܧ .
Planes ܨܪܦܤ and ܥܨܣ are intersecting at line
ܮܨ .Suppose that ܭܤ is the height line of
passesthrough ܭܤ .ܮܨ ⊥ ܭܤ then, ܨܮܤ triangle
ܥܨܣ and is perpendicular to ܮܨ on ܥܨܣ ,hence
ܭܤ is the distance between ܤ to ܥܨܣ .
Look at ∆ܮܤܨ
Figure 1.16
= ܤܮ ,݉ܿ 6 = ܨܤ
ଵ
ଶ
∙ 6√2 = 3√2 ܿ݉
ܮܨଶ = ܤܮଶ + ܨܤଶ = 18 + 36 = 54
ܮܨ√ = 54 = 3√6 ܿ݉
= ߠ sin
ி
ி
=
ଷ√
=
ଵ
ଷ
√6
= ߠ sin
ߠ sin ܮܤ = ܭܤ ⟹
= 3√2 ቀ
ଵ
ଷ
√6ቁ
= √12 = 2√3
Hence, the distance between point ܤ to plane ܥܨܣ is 2√3 cm.
4. The Distance between Two Parallel Lines
Consider Figure 1.17. assume that two lines ݃ and ℎ are parallel to each other
and are located on plane ߙ .Suppose that line ݈ is perpendicular to both lines
݃ and ℎ and also intersects ݃ and ℎ at points ܲ and ܲ′, respectively. The
distance between lines ݃ and ℎ is the legth of line segment ܲܲ′.
Figure 1.17
5. The Distance between Two Skew Lines
Two lines ݃ and ℎ are said to be skew to each other if they are neither parallel
nor located on a same plane.
Look at Figure 1.18. line ܧܣ is skew to line ܪܩ .No matter how long we
extend these lines, they will never intersect each other, as do lines ܪܧ and ܨܤ
and lines ܪܧ and ܩܤ .
Following are a set of steps to find the distance between two skew lines.
Mathematics for Grade XII 16
Observe Figure 1.19
Figure 1.18 Figure 1.19
1. Assume that lines ݃ and ℎ are skew to each other. Draw a line ݃′ that is
parallel to ݃ and intersects ℎ.
2. Draw a plane ߙ that contains lines ݃′ and ℎ.
3. Draw a plane ߚ that is perpendicular to plane ߙ and contains line ݃. Plane
ߚ intersects line ℎ at ܲ.
4. Draw a line that passes through ܲ and is perpendicular to ݃ and suppose
that this line intersects ݃ at ܳ.
5. ܲܳ is the distance between lines ݃ and ℎ.
If lines ݃ and ℎ are skew at a right angle, then the distance between ݃ and ℎ
can be determined as follows (Figure 1.20).
1. Draw a plane ߙ that contains line ݃ and is perpendicular to line ℎ.
2. Assume that line ℎ passes through plane ߙ at point ܲ.
3. Draw a line that passes through ܲ and is perpendicular to ݃ and assume
that this line intersects ݃ at point ܳ.
4. ܲܳ is the distance between lines ݃ and ℎ that are skew at a right angle.
Figure 1.20
Mathematics for Grade XII 17
Let’s, Determine the Distance Between Two Skew Lines!
Known T.ABCD is a limas as follows.
Determine the distance between the line AD and line TB.
Answer:
Lines AD and TB skew each other. The distance between lines AD and TB can be
determined by the following steps.
1. Determine a plane through one of the line and parallel to the other lines.
Suppose the plane TBC through line TB and is parallel to the line AD. The
distance between the lines AD and TB is the same as the distance between
AD and plane TB.
2. Determine the distance between line AD and plane TBC that parallel to
AD. To do this, select an arbitrary point on the AD and then determine the
distance of that point to plane TBC. Suppose that point P is choosen,
namely the midpoint of AD. The distance between AD and plane TBC is
the same as the distance between the point P and plane TBC.
3. Determine the distance between point P and plane TBC. How? Determine
the point R on the plane TBC so that the PR is perpendicular to the plane
TBC. Suppose point Q is the midpoint of BC, then the line PQ is
perpendicular to the line BC. In general, the line connecting the point P
with the line TQ is perpendicular to the line on the plane TBC which is
parallel to BC. This means that the distance between the point P and the
plane TBC is the same as the distance between the point P and the line TQ.
So it can be written as follows.
Mathematics for Grade XII 18
The distance (AD, TB) = The distance (AD, ...)
= The distance (... TB)
= The distance (P, ...)
= The distance (P, ...)
The TBC triangle is isosceles with TB = TC and Q is the midpoint of BC,
then the triangle TBQ is right-angled at Q so that:
TB = TC = ⋯ cm
BQ =
1
2
BC =
1
2
× 10 = ⋯ cm
TQ = ඥTBଶ − BQଶ
= ඥ… − 5ଶ
= √169 − ⋯
= √…
= ⋯ cm
Look at the TPQ triangle with TP = TQ = ⋯ cm and PQ = ⋯ cm
Mathematics for Grade XII 19
There are two ways to determine the length of PR.
Method 1: Use the triangle area formula
s =
1
2
(TP + TQ + PQ)
=
1
2
(12 + ⋯ + 10)
=
1
2
× …
= ⋯ cm
The area of ∆ܶܲܳ:
L = ඥs(s − TP)(s − TQ)(s − PQ)
= ඥ17(17 − ⋯ )(17 − 12)(17 − ⋯ )
= √17 × 5 × … × 7
= √25 × …
= ⋯ cmଶ
The area of ∆TPQ = ଵ
ଶ
× TQ × PR, so:
1
2
× TQ × PR = 5√119
⇔
1
2
× … × PR = 5√119
Mathematics for Grade XII 20
⇔ ⋯ × PR = 5√119
⇔ PR = ⋯ √119
Obtained the length of PR = ⋯ cm.
Method 2: Using the Pythagorean formula
Suppose the length of QR = x cm, then length TR = (12-x) cm.
∆PQR right angle at R, then:
PRଶ = PQଶ − QRଶ
= ⋯ − xଶ
= ⋯ − xଶ
...(i)
∆PRT right angle at R, then:
PRଶ
= PTଶ − TRଶ
= ⋯ − (12 − x)ଶ
= 144 − (… − 24x + xଶ)
= ⋯ − 144 + ⋯ − xଶ
= ⋯ − xଶ
...(ii)
From equations (i) and (ii) we get:
ݔ − 100
ݔ − ݔ24 = ଶ
ଶ
⋯ = ݔ24⇔
⋯ = ݔ ⇔
= ݔ Substitute
ଶହ
to the equation (i) and obtained:
PRଶ = 100 − xଶ
= ⋯ − ൬
25
6
൰
ଶ
Mathematics for Grade XII 21
=
3.600
36 − ⋯
= ⋯
PR = ට
ଶ.ଽହ
ଷ
= ට
ଶହ×…
ଷ
=
…
√119 cm
Obtained length PR = ⋯ cm.
Thus, the distance between the AD line and the TB line is ହ
√119 cm.
6. The Distance between a Parallel Line and a Plane
Look at Figure 1.21. Line ݃ is parallel to plane ߙ .Draw a line that passes
through an arbitary point ܲ on line ݃ and is perpendicular to plane ߙ ,this line
passes through plane ߙ at point ܲ′. The distance between line ݃ and plane ߙ is
ܲܲ′.
The distance between line ݃ and plane ߙ which are parallel to each other is the
leght of line segment ܲܲ′, where ܲ is an arbitary point on line ݃ and ܲ′ is the
projection of point ܲ on plane ߙ .
Figure 1.21
7. The Distance between Two Parallel Planes
Look at Figure 1.22. Suppose that planes ߙ and ߚ are parallel to each other.
Choose any arbitary point on ߚ ,say point ܲ. Suppose that line ݃ passes through
ܲ and is perpendicular to plane ߙ ,and point ܳ is the pass-through point of line
݃ on plane ߙ ܳܲ .is the distance between the two planes ߙ and ߚ .
the distance between the two planes ߙ and ߚ which are parallel to each other is
the length of line segment ܲܳ, where ܲ is an arbitary point on plane ߚ and ܳ
is the projection of point ܲ on plane ߙ .
Mathematics for Grade XII 22
Figure 1.22
Example 5
ܦܥܤܣ .ܪܩܨܧ has a length of 8 cm, a width of 4 cm, and a height of 6 cm.
Find the distance between:
ܪܩ and ܤܣ .a
b. ܪܣ and plane ܨܩܥܤ
c. Planes ܨܩܥܤ and ܧܪܦܣ
d. Lines ܧܣ and ܪܥ
Solution
a. ܤܣ and ܪܩ are located on plane ܪܩܤܣ .ܤܣ and ܪܩ are parallel lines,
and so the distance between ܤܣ and ܪܩ can be represented by the length
.ܩܤ of
6 = √ܩܤ
ଶ + 4ଶ = 2√13 = 7.21 cm
∴ The distance between ܤܣ and ܪܩ is 7.21 cm.
b. ܪܣ is parallel to plane ܨܩܥܤ .ܪܣ is parallel to ܩܤ on plane ܨܩܥܤ .The
distance between ܪܣ and ܨܩܥܤ can be represented by the length of
.cm 8 = ܤܣ ,ܤܣ
∴ The distance between ܪܣ and plane ܨܩܥܤ is 8 cm.
c. ܨܩܥܤ ⧵⧵ ܧܪܦܣ .The line segment ܤܣ can be used to represent the
distance between these two planes, since ܤܣ ⊥ ܨܩܥܤ .
∴ The distance between planes ܨܩܥܤ and ܧܪܦܣ is 8 cm.
d. Line segments ܧܣ and ܪܥ are skewed. Line ܪܦ is parallel to ܧܣ and
intersects ܪܥ at point ܪ .Lines ܪܦ and ܪܥ form the plane ܪܩܥܦ .Line
ܧܪ is perpendicular to plane ܪܩܥܦ and intersects line ܧܣ also
perpendicularly, so that ܧܪ can represent the distance between ܧܣ and
.ܪܥ
∴ The distance between planes ܧܣ and ܪܥ is 4 cm.
Mathematics for Grade XII 23
Examples
1. It is known that the KLMN.OPQR is a box with KL = LM = 6 cm and LP =
9 cm. Determine the distance of the following point pairs.
a. Point O and point M.
b. Point P and point S if S at ON with OS: SN = 1: 2.
Answer:
The KLMN.OPQR is a box and the MNOP is the diagonal plane that are
described as follows.
a. The distance between point O and point M is the same as the length of OM
line segment.
∆OKN right angle at K with OK = 9 cm and KN = 6 cm, then:
ON = ඥOKଶ + KNଶ
= ඥ9
ଶ + 6ଶ
= √81 + 36
= √117
= 3√13
∆ONM right angle at N with ON = 3√13 cm and NM = 6 cm, then:
OM = ඥONଶ + NMଶ
= ට(3√13)
ଶ + 6ଶ
= √117 + 36
= √153
= 3√17
Mathematics for Grade XII 24
So, the distance between point O and point M is 3√17 cm.
b. The distance between point P and point S is equal to length of line segment
PS.
PO = 6 cm
OS: SN = 1: 2, then:
OS =
1
1 + 2 ON
=
1
3
× 3√13
= √13 cm.
∆ܱܲܵ right angle at O, then:
PS = ඥPOଶ + OSଶ
= ට6
ଶ + ൫√13൯
ଶ
= √36 + 13
= √49
= 7 cm.
So, the distance between point P and point S is 7 cm.
2. In the T.PQRS pyramid, the length of PQ = 8 cm, PS = 6 cm, and height of
the pyramid 12 cm. Determine the distance between:
a. Point P and line QR
b. Point P and line TR
c. Lines PQ and SR
Answer:
Limas T.PQRS is described as follows.
Mathematics for Grade XII 25
a. The distance (P, QR) = The distance (P, Q)
= PQ
= 8 cm.
So, the distance between point P and line QR is 8 cm.
b. ∆PQR right angle at Q with PQ = 8 cm and QR = 6 cm, then:
PR = ඥPQଶ + QRଶ
= ඥ8
ଶ + 6ଶ
= √64 + 36
= √100
= 10 cm
Point O is the midpoint of PR, then OR = ଵ
ଶ
PR = ଵ
ଶ
× 10 = 5 cm.
∆ܱܴܶ right angle at O with TO = 12 cm and OR = 5 cm, then:
TR = ඥTQଶ + QRଶ
= √12ଶ + 5ଶ
= √144 + 25
= √169
= 13 cm
Look at the following ∆TPR picture.
Mathematics for Grade XII 26
The distance (P, TR) = The distance (P, D)
= PD
The area of ∆TPR can be written as L = ଵ
ଶ
× TR × PD or L = ଵ
ଶ
× PR ×
TO, then:
1
2
× TR × PD =
1
2
× PR × TO
⇔
1
2
× 13 × PD =
1
2
× 10 × 12
⇔ PD =
1
2
× 10 × 12
1
2
× 13
⇔ PD =
120
13
So, the distance between point P and line TR is ଵଶ
ଵଷ
cm.
c. The distance (PQ, SR) = The distance (P, SR)
= The distance (P, S)
= PS
= 6 cm.
So, the distance between the line PQ and the line SR is 6 cm.
3. Consider the following picture.
The lines AB, BC, and BD are perpendicular to B. The length of AB =
6√3cm and BC = BD = 6 cm. Determine the distance between the
following points and plane.
a. Point A and plane BCD
b. Point B and plane ACD
Mathematics for Grade XII 27
Answer:
a. The line AB is perpendicular to the lines BC and BD, then the line AB
is perpendicular to the plane BCD, so that:
The distance (A, BCD) = The distance (A, B)
= AB
= 6√3 cm
b. Look at the following picture.
Suppose E is the midpoint of the CD, then
The distance (B, ACD) = distance (B, AE)
= distance (B, F)
= BF
∆BCD right angle at B with BC = BD = 6 cm, then:
CD = ඥBCଶ + BDଶ
= ඥ6
ଶ + 6ଶ
= √36 + 36
= √72
= 6√2
CE =
1
2
CD =
1
2
× 6√2 = 3√2 cm
∆ܧܥܤ right angle at E with BC = 6 cm and CE = 3√2 cm, then:
BE = ඥBCଶ − CEଶ
= ට6
ଶ − ൫3√2 ൯
ଶ
Mathematics for Grade XII 28
= √36 − 18
= √18
= 3√2 cm.
∆ܧܤܣ right angle at B with AB = 6√3 cm and BE = 3√2 cm, then:
AE = ඥABଶ + BEଶ
= ට൫6√3 ൯
ଶ
+ ൫3√2 ൯
ଶ
= √108 + 18
= √126
= 3√14
The area of ∆ܧܤܣ can be written as L = ଵ
ଶ
× AE × BF or L = ଵ
ଶ
× BE ×
AB, then:
1
2
× AE × BF =
1
2
× BE × AB
⇔
1
2
× 3√14 × BF =
1
2
× 3√2 × 6√3
⇔ BF =
1
2
× 3√2 × 6√3
1
2
× 3√14
⇔ BF =
6√3
√7
×
√7
√7
⇔ BF =
6
7
√21 cm
So, the distance between point B and plane ACD is
√21 cm.
Mathematics for Grade XII 29
4. It is known that PQRS. TUVW is a box with the length of PQ = 6 cm, QR
= 4 cm, and RV = 3 cm. Determine the distance between:
a. Line PQ and plane SRVW;
b. Line SW and plane PRVT;
c. Planes PSWT and QRVU;
d. Planes PUW and QSV.
Answer:
a. The PQRS.TUVW is a box that described as follows.
The distance (PQ, SRVW) = The distance (P, SRVW)
= distance (P, S)
= PS
= 4 cm
So, the distance between the line PQ and the plane SRVW is 4 cm.
b. Look at the picture of the top side of the box below.
The distance (SW, PRVT) = The distance (W, PRVT)
= distance (W, VT)
= distance (W, X)
= WX
Mathematics for Grade XII 30
∆TUV right angle at U, then:
TV = ඥTUଶ + UVଶ
= ඥ6
ଶ + 4ଶ
= √36 + 16
= √52
= 2√13 cm.
The area of TVW can be written as ܮ=
ଵ
ଶ
= ܮ or× ܸܶ × ܹܺ
ଵ
ଶ
× ܹܶ ×
ܹܸ, then:
1
2
× TV × WX =
1
2
× TW × WV
⇔
1
2
× 2√13 × WX =
1
2
× 4 × 6
⇔ √13 WX = 12
⇔ WX =
12
√13
×
√13
√13
⇔ WX =
12
13 √13 cm
So, the distance between line SW and plane PRVT is ଵଶ
ଵଷ
√13 cm.
c. The distance (PSWT, QRVU) = the distance (P, QRVU)
= distance (P, Q)
= PQ
= 6 cm.
So, the distance between the plane PSWT and the plane QRVU is 6 cm.
d. Look at the following picture.
The distance (PUW, QSV) = The distance (K, QSV)
= distance (K, LV)
= distance (K, M)
= KM
∆LKV right-angle at K with LK = 3cm and KV = √13 cm, then:
LV = ඥLKଶ + KVଶ
Mathematics for Grade XII 31
= ට3
ଶ + ൫√13൯
ଶ
= √9 + 13
= √22 cm
The area of ∆ܭܮ ܸcan be written as L = ଵ
ଶ
× LV × KM or L = ଵ
ଶ
× LK ×
KV, then:
1
2
× LV × KM =
1
2
× LK × KV
⇔
1
2
× √22 × KM =
1
2
× 3 × √13
⇔ KM =
1
2
× 3 × √13
1
2
× √22
⇔ KM =
3√13
√22
×
√22
√22
⇔ KM = ଷ
ଶଶ
√286 cm
So, the distance between the plane PUW and plane QSV is ଷ
ଶଶ
√286 cm.
5. Known ABCD.EFGH is a cube with the length of the side is 8 cm.
Determine the distance between the following two lines.
a. BG and DH
b. EG and CH.
Answer:
a. Look at the following picture.
Mathematics for Grade XII 32
Lines BG and DH are skew each other. The plane BCGF through the
line BG and is parallel to the line DH, then:
The distance (BG, DH) = distance (BCGF, DH)
= distance (BCGF, D)
= distance (C, D)
= CD
= 8 cm
So, the distance between the lines BG and DH is 8 cm.
b. Look at the following picture.
Lines EG and CH are skew each other. The plane that pass through CH
and parallel to EG is plane ACH, then:
The distance (EG, CH) = distance (EG, ACH)
= distance (Q, ACH)
= distance (Q, PH)
= distance (Q, R)
= QR
HF is a diagonal plane, hence HF= 8√2 cm.
HQ =
1
2
HF =
1
2
× 8√2 = 4√2 cm
∆ܪ ܳܲright-angle at Q, then:
HP = ඥHQଶ + QPଶ
= ට൫4√2 ൯
ଶ
+ 8ଶ
= √32 + 64
= √96
= 4√6 cm
Mathematics for Grade XII 33
The area of ∆ܣ ܳܲcan be written as ܮ=
ଵ
ଶ
= ܮ orܲ × ܴܳ ܪ ×
ଵ
ଶ
×
,Soܳ × ܲܳ. ܪ
1
2
× HP × QR =
1
2
× HQ × PQ
⇔
1
2
× 4√6 × QR =
1
2
× 4√2 × 8
⇔ 2√6 QR = 16√2
⇔ QR =
16√2
2√6
=
8
√3
=
8
3
√3 cm
So, the distance between line EG and line CH is ଼
ଷ
√3 cm.
6. Given the cube ABCD.EFGH with side 2√2 cm. If point P is in the middle
of AB and point Q is in the middle of BC, then the distance between point
H and line PQ is ... cm.
A. √15 D. 3√2
B. 4 E. √19
C. √17
Solution:
Pay attention to triangle APH
AP = √2, AH = 4, so
HP = ට4
ଶ + (√2)
ଶ = √18
Pay attention to triangle PBQ with ܲܳ = 2
Mathematics for Grade XII 34
Then, Pay attention to triangle PQH
OH = √18 − 1 = √17
The distance between point H and line PQ = ܱܪ√ = 17cm (C)
7. On the cube ABCD. EFGH, point P lies on segment BG so 3×PG=2×BP.
The point Q is the point of intersection of the line HP and the plane ABCD.
If the side of the cube is 6 cm, the area of triangle APQ is ... cmଶ
.
a. 9√2 d. 27√2
b. 12√2 e. 36√2
c. 18√2
Solution:
BP: PG = 3: 2
BP =
3
5
BG =
3
5
. 6√2
Mathematics for Grade XII 35
Congeniality ∆AQH with ∆BQP:
BQ
AQ =
BP
AH
⟺
6 + x
12 + x =
3
5
. 6√2
6√2
⟺ 30 + 5x = 36 + 3x
2x = 6
x = 3
Get AQ = 12 + 3 + 15 cm.
Since BP is perpendicular to AQ, we get:
L∆AQP = ଵ
ଶ
. AQ. BP = ଵ
ଶ
. 15. ଷ
ହ
. 6√2 = 27√2 (D)
8. Cube ABCD.EFGH has a side length of 4 cm. Point P is the middle of EH.
The distance from point P to line BG is ... .
Solution:
Pay attention to the triangle BPG.
O is the projection of point P on line BG. The distance from point P to
line BG is the length of PO, where
GO =
1
2
× plane diagonal BCFG =
1
2
× 4√2 = 2√2
PG = ඥ(GH)ଶ + (PH)ଶ = ඥ4
ଶ + 2ଶ = √20 = 2√5
So, PO = ඥ(PG)ଶ − (GO)
ଶ = ට(2√5)
ଶ − ൫2√2൯
ଶ = √20 − 8 = √12 =
2√3 cm.
Mathematics for Grade XII 36
9. Cube ABCD.EFGH has a side length of 8 cm. The distance from point E
to the DGD plane is ... cm.
Solution:
The distance from point E to the BGD plane is EP.
EC = 8√3 (space diagonal)
EP =
2
3
EC =
2
3
. 8√3 =
16
3
√3
10. Cube ABCD.EFGH has a side length of 6√3 cm. The distance between
ACH and EGB is ... cm.
Solution:
space diagonal length = 6√3 × √3 = 18 cm
The distance between ACH and EGB =
ଵ
ଷ
× space diagonal = ଵ
ଷ
× 18 =
6 cm.
Mathematics for Grade XII 37
1. Given a cube ܦܥܤܣ .ܪܩܨܧ having an edge length of 4 cm. Supoose that ܲ
is the midpoint of edge ܩܪ and ܯ is the midpoint of edge ܪܧ .Calculate the
distance between:
a. ܣ and ܲ
b. ܲ and line ܥܣ
c. ܨ and plane ܪܥܣ
d. Lines ܤܧ and ܪܥ
e. ܯ and line ܥܣ
f. ܧܣ and plane ܨܪܦܤ
g. Planes ܩܦܤ and ܪܨܣ
h. ܨ and line ܪܣ
i. ܲ and line ܥܤ
ܨܦ and ܯܲ .j
2. Given ܶ. ܥܤܣ is a regular tetrahedron with edge-length of 6 cm, find the
distance between point ܶ and plane ܥܤܣ
Classroom Activities 2
Mathematics for Grade XII 38
CHAPTER REVIEW EXERCISE
A. Choose the right answer.
1. Look at the following picture. If ADHE is square and T.BCGF is a pyramid
with the height of the side is 2√6 cm, the distance between point A and
point T is ... cm.
A. 12 D. 9√2
B. 2√38 E. 14√2
C. 4√10
2. Known ABCD.EFGH is a cube with the length of the side is 8 cm. The
distance between point C to the diagonal of AH is ... cm.
A. 4√6 D. 2√6
B. 4√3 E. 2√3
C. 4√2
3. It is known T.ABCD pyramid which the length of the base-side is 8 cm
and the length of the side is 8√2 cm. The distance between point C and
line TA is ... cm.
A. 8√3 D. 4√3
B. 8√2 E. 4√2
C. 4√6
4. It is known that KLMN.PQRS is a box with KL = 3 cm, LM = 4 cm, and
KP = 12 cm. The distance from point R to the line PM is ... cm.
A. ଷହ
ଵଷ
D. ହ
ଵଷ
B. ସ
ଵଷ
E.
ଵଷ
C. ସହ
ଵଷ
5. In the T.ABCD pyramid the length of AB = 6 cm, AD = 8 cm, and TA =
10 cm. The distance of the point T to the plane ABCD is ... cm.
A. 5√3 D. 5√6
B. 10 E. 15
C. 5√5
Mathematics for Grade XII 39
6. Look at the picture. If TK, KL, and KM are perpendicular to K and length
of TK = 2√2 cm and KL = KM = 4 cm, the distance from point K to the
plane TLM is ... cm.
A. 2√3 D. √3
B. 2√2 E. √2
C. 2
7. On the ABCD. EFGH cube, the length of the side is 12 cm. The distance
of point E to the plane BGD is ... cm.
A. 12√3 D. 9√2
B. 9√3 E. 8√2
C. 8√3
8. The ABCD.EFGH is a cube which the length of the side is 9 cm. Point P
is the intersection between AH and DE and point Q is the midpoint of GH.
The distance between the lines PQ and AG is ... cm.
A. 3√6 D. ଶ
ଷ
√6
B. 2√6 E. ଵ
ଷ
√6
C. ଷ
ଶ
√6
9. It is known that PQRS.TUVW is box with the length of PQ = 8 cm, QR =
6 cm, and RV = 4 cm. The distance of the line RV to the plane QSWU is
... cm.
A. 4,8 D. 6,4
B. 5,4 E. 7,2
C. 6,0
10. The KLMN.OPQR is a box with KL = 14 cm, KN = 10 cm, and KO = 8
cm. The distance between the line KR and MQ is ... cm.
A. 16 D. 12
B. 15 E. 10
C. 14
Mathematics for Grade XII 40
B. Work on the following questions.
1. It is known that the ABCD.EFGH is a box with AB = 8 cm, BC = 6 cm,
and AE = 4√6 cm. Determine the distance between:
a. Point A to the midpoint of the plane EFGH;
b. Point A to the plane EFGH.
2. Known ABCD. EFGH is a cube with length 6 cm. If P is the midpoint of
GH, determine:
a. The distance of point P to line CF;
b. The distance of Point P to the plane ACGE.
3. It is known that the PQRS.TUVW is a box with a length of PQ = 15 cm
and QR = 9 cm. It is known that the surface area of the box is 846 cmଶ
,
determine:
a. The distance between the lines PQ and WV;
b. The distance between the line PQ and the plane RSTU.
4. Known ABCD. EFGH is a cube with the length of the side is 6 cm.
Determine the distance between the planes ACH and BEG.
5. There is known T.ABCD limas with the height of the side is 8 cm, AB =
4√3 cm, and BC = 6 cm. Determine the distance between the lines AD and
TB.
6. Given a regular rectangular pyramid ܶ. ܦܥܤܣ with ܤܣ = 6 cm and ܶܣ=
5 cm, calculate the distance between:
a. ܶ and edge ܤܣ ,
b. ܶ and base plane,
c. ܧ and plane ܶܥܤ ,if ܧ is the midpoint of ܦܣ
7. Consider a cube ܦܥܤܣ .ܪܩܨܧ with an edge length = 4 cm.
i. Draw and calculate the distance between:
a. line ܪܨ and plane ߙ that contains ܥܣ and is parallel to ܪܨ ,
b. line ܩܧ and a line that passes through ܤ and skew to ܩܧ ,
c. planes ܩܧܤ and ܪܥܣ
ii. Prove that the distance between ܨ and ܩܧܤ=
ଵ
ଷ
ܦܨ ,and the distance
= ܪܥܣ and ܨ between
ଶ
ଷ
.ܦܨ
8. Consider a cube ܦܥܤܣ .ܪܩܨܧ ,ܤܣ = 6 cm, and ܲ as the midpoint of ܪܩ .
Point ܳ is on ܦܣ such that 2ܣ = ܳܦ ܳand ܴ is the midpoint of ܤܣ .
Calculate and draw the distance:
ܳܤ .a
b. ܲܳ
ܨܤ and ܣ .c
d. ܲ and ܤܧ
ܧܦܤ and ܣ .e
ܪܨܥ and ܧܦܤ .f
9. Consider a regular triangular pyramid ܶ. ܦܥܤܣ where ܶܦ is at a right
angle with the base plane. The edges ܤܣ = ܥܤ = ܦܥ = ܦܣ = 4 cm and
ܶܦ = 4√3 cm. Find the distance between point ܣ and the plane ܶܥܤ .
Mathematics for Grade XII 41
10. Consider a tetrahedron ܦ .ܥܤܣ with ܤܣ = ܥܣ = 9 cm, ∠ܦܥܤ
isequlateral, and the distance between point ܦ and plane ܥܤܣ is ܽ cm.
a. Prove that ܦܣ and ܥܤ are perpendicularly skew to each other.
b. Calculate the distance between point ܣ and plane ܦܥܤ .
c. Calculate the distance between ܦܣ and ܥܤ .
11. On a cube ABCD.EFGH, N is the intersection point of diagonals of the top
face and K is the intersection point of CE and AN.
a. Find the ratio of EK to KC
b. If K’ is the projection of K on ABCD, prove that AK’ : K’C = EK :
KC = 1: 2
Mathematics for Grade XII 42
Angles in Space
On a planar surface, an angle can only be formed bt two non-parallel lines. In space
(three-dimensional), the concept of angles can be extended to include an angle
between two intersecting lines, an angle between two skew lines, an angle between a
line and plane, and an angle between two planes.
1. An Angle between Two Lines
Consider Figure 1. 23. Line g and line h are
both located on a same plane and intersect
each other at point O. An angle formed by
Figure 1. 23
these lines g and h, written (݃, ℎ), is ∠ܱܳܲ or ∠ܳ′ܱܲ′. If the two lines are skew
to each other, then each of them is licated on a different plane. We can find an
angle between two skew lines by shifting one of them (or both) such that the two
lines will be located on the same plane. This way, both lines will intersect with
each other. An angle formed after this shifting is the desired angle between the
two lines.
As an example, in cube ABCD.EFGH (Figure 1.24), the
angle between lines FH and AD is the same as the angle
between lines BD and AD (found by shifting FH), that
is, ∠ܣܤܦ is 45o
. We can also shift AD so that we have
∠ܧܪܨ which is 45o
.
Between two skew lines, we can find an angle as follows:
(see Figure 1.25).
Figure 1.24
1. Assume that a line h is
located on plane a and a line g is
located outside of a. The two lines
g and h are skew to each other.
2. Determine one arbitrary
point P in space.
3. Draw a straight line passing
through P and parallel to line ݃ (name it ݃′),
then draw another line passing through P and
parallel to h (name it h’).
4. ∠(݃, ℎ) = ∠(݃′, ℎ′) = ߠ
Figure 1.25
Mathematics for Grade XII 43
We can also arbitrarily choose a point P located on line ݃ (this way, we just
have to find a line ℎ′ that passes through P and is parallel to h). In this case,
∠(݃, ℎ) = ∠(݃, ℎ′) (Figure 1.26 (a)). If we select P as located on line ℎ, then
we just have to find a line ݃′ passing through P and parallel to ݃ so that ∠(݃, ℎ)
= ∠(݃′, ℎ) (Figure 1.26 (b)).
Figure 1.26
Example 6
Look at the cube ABCD.EFGH below. Find the angles that are formed between
each following pair of lines:
c. AE and ED
d. AH and HC
e. AD and BG
f. EC and HD
Solution
a. Lines AE and ED intersect each other at E. The
angle between AE and ED is ∠ܦܧܣ .Observe
that ∆ ܦܧܣ is an isosceles right triangle.
So, ∠ܦܧܣ = 45o
Lines AE and ED form a 45o
-angle.
Mathematics for Grade XII 44
b. AH and HC intersect each other at H and are located on the same plane
AHC. The angle between AH and HC is ∠ܥܪܣ .
ܥܪܣ ∆ Observe
AH=HC=CA=4 cm.
Hence AHC is equilateral
So ∠ܥܪܣ = 60o
Lines AH and HC form a 60o
-angle.
c. AD and BG are skew to each other. Line BG is
parallel to line AH located on plane ADHE. ∠(ܦܣ ,ܩܤ)∠ = (ܦܣ ,ܪܣ = (
45o) =ܪܣܦ)∠
.
d. EC and HD are skew to each other. HD is parallel to line GC located on
.ܩܥܧ ∆ Observe). ܩܥܧ)∠ = ܥܩ , ܧ)∠ = ܦܪ ,ܥܧ)∠ .ACGE plane
Classroom Activities 3 1. In a cube ABCD.EFGH, find the angle between:
a. CD and FH
b. BD and FG
c. AF and BD
d. FG and AD
e. CF and AD
f. FG and AH
g. FC and BG
h. AH and BF
i. BE and CF
2. Determine in a cube ABCD.EFGH the number of lines that form a 45o - angle with AC
Mathematics for Grade XII 45
2. An Angle between a Line and a Plane
If a line ݃ is not perpendicular to a plane ߙ ,then
the angle between line ݃ and plane ߙ is an acute
angle that is formed by line ݃ and its projection on
ߙ݃)ᇱ), or ∠(݃, ߙ .(′݃ ,݃)∠ = (In other words, if ݃
and ߙ are parallel to each other, then ∠(݃, ߙ= (
0°). If ݃ is not parallel to ߙ ,then ∠(݃, ߙ= (
∠(݃, ݃′) where ݃′ is the projection of ݃ on plane
ߙ ,and 0 ൏ ∠(݃, ݃′) ≤ 90°. Consider Figure 1.27,
.ߠ = (′݃ ,݃)∠ = (ߙ ,݃)∠
Example 7
A cube ABCD.EFGH has an edge length of 6 cm. Find ∠(ܪܣ ,ܨܪܦܤ .(
Solution
AM ⊥ BDHF. The projection of line AH on
plane BDHF is MH.
ܯܪܣ∠ =(ܪܯ ,ܪܣ)∠ = (ܨܪܦܤ ,ܪܣ)∠
ܪܣ = 6√3 cm, ܪܯ 3√6 cm,
ܯܣ 3√2 cm,
(ܯܣ)
(ܯܪ) + ଶ
ଶ = ൫3√2൯
ଶ + ൫3√6 ൯
ଶ
= 18 + 54
= 72
(ܪܣ) =
ଶ
∆ܯܪܣ is a right triangle at M. ∠ܯܪܣ = ߠ
= ߠ tan
3√2
3√6
=
1
3
√3
ߠ = tanିଵ 1
3
√3 = 30°
Figure 1.27
Mathematics for Grade XII 46
Figure 1.28
3. An Angle between Two Planes
An angle between two planes ߙ and
ߚ is written or ∠(ߙ ,ߚ .(If ߙ and ߚ is
parallel to each other, then ∠(ߙ ,ߚ= (
0°. If ߙ and ߚ are not parallel, then ߙ
and ߚ will intersect each other on the
intersection line ݃ (Figure 1.28). A
resting plane is a plane that is drawn
at a right angle to the intersection line
of intersecting planes. In Figure 1.28,
the resting plane ߛ is perpendicular to line ݃(ߙ ,ߚ .(The included angle
between lines of intersection of resting plane and the two planes ߙ and ߚ is
called resting angle (ߠ .(In Figure 1.28 we can see that ݃ଵ = intersection line
(ߛ ,ߙ (and ݃ଶ = intersection line (ߛ ,ߚ (are perpendicular to ݃ = intersection
line (ߙ ,ߚ .(The angle between the two planes can be determined as follows:
1. Choose an arbitrary point on line ݃, say the point P.
2. Draw a line on plane ߙ that is perpendicular to line ݃ and passes through
point P, name the line ݃ଵ.
3. Draw a line on plane ߚ that is perpendicular to line ݃ and passes through
point P, name the line ݃ଶ.
4. The plane formed by ݃ଵ and ݃ଶ is the resting plane ߛ and the angle
between ݃ଵ and ݃ଶ is the resting angle (ߠ(
.ߠ = (ଶ, ݃ଵ) = ∠ (݃ߚ ,ߙ)∠ ,Hence
Example 8
Consider a cube ABCD.EFGH having an edge length of 4 cm. Find the angle
included between plane BDG and ABCD.
Solution
Plane BDG and ABCD intersect each other
at the intersection line BD. Choose a point P
that is the midpoint of BD.
BD ⊥ PC (diagonals of square ABCD)
BD ⊥ PG (height or bisect of equilateral
∆BDG)
Mathematics for Grade XII 47
ܩܲܥ∠ = (ܦܥܤܣ ,ܩܦܤ)∠
ߠ = ܩܲܥ∠
= ߠ tan
ସ
ଶ√ଶ
= √2
ߠ = tanିଵ √2
ߠ = 54.7°
°7.54) = ܦܥܤܣ ,ܩܦܤ)∠ ∴
1. On cube ABCD.EFGH, point P is the midpoint of edge FG. Find:
(ܩܦ,ܲܣ)∠ .a
(ܤܣ ,ܲܥ)∠ .b
(ܪܣ ,ܲܤ)∠ .c
(ܥܤ ,ܲܦ)∠ .d
(ܨܪ ,ܲܣ)∠ .e
(ܦܣ ,ܲܧ)∠ .f
2. A regular rectangular pyramid T.PQRS has an edge leght of its base plane
equals 4√2 cm. Given TP = 8 cm, determine and calculate:
a. ∠(ܶܲ,ܳܵ)
b. ∠(ܶܲ, ܴܵ)
3. Consider a cube KLMN.PQRS with edge length of 6 cm and point A as
midpoint of KL. Calculate:
(ܰܭ,ܣܵ)∠ cos. a
b. sin ∠(ܳܣ(ܲܰ ,
c. tan ∠(ܴܣ(ܳܲ ,
4. Consider a cube cube ABCD.EFGH with edge length of 8 cm. Determine
and calculated:
(ܪܥܣ ,ܨܪ)∠ cos. a
(ܧܦܤ ,ܧܥ)∠ sin. b
(ܪܨܥ ,ܪܣ)∠ tan. c
(ܩܦܤ ,ܩܥ)∠ tan. d
5. On cube ABCD.EFGH, point P is the midpoint of edge EH. Calculate:
(ܲܧ ,ܤܪ)∠ sin. a
(ܨܤ ,ܲܥ)∠ tan. b
(ܨܤ ,ܲܧ)∠ cos. c
(ܨܦ,ܲܪ)∠ sin. d
Mathematics for Grade XII 48
6. A flagpole PT standing upright on the P corner of a rectangular field PQRS
which side-lenghts are PQ= 160 m and QR = 120 m. The elevation angle
of point T as measured from Q is 12.5o
. Calculate the elevation angle of
point T as measured from R from S.
7. Consider a regular rectangular pyramid T.ABCD with base-plane’s edgelength of 6 cm and lateral edge of 8 cm. Find the angle between:
a. TA and plane ABCD
b. TA and plane TBC
8. Consider a tetrahedron A.PQR having a right triangle base plane where PQ
= PR, and ܣ = ܲ5√3 cm and perpendicular to base plane. If QR = 10 cm,
calculate tan ∠ (AQR,base plane).
9. Consider a cube ABCD.EFGH having an edge length of 6 cm. Suppose
that K is located on edge FG such that FK : KG = 3 : 2. Find
.(ܥܦܤ ,ܤܦܭ)∠ tan
10. Consider a cube ABCD.EFGH having an edge length of 6 cm. Calculate
.(ܧܦܤ ,ܩܦܤ)∠ sin
11. Three edges that meet (intersect) at point A of a pyramid T.ABC are all
orthogonal (perpendicular to each other). The length of AB = AC =4√2 cm
and AT= 4√3 cm. Calculate:
(ܥܤܣ ,ܶܥܤ)∠ .a
(ܥܤܣ ,ܶܥܤ)∠ tan. b
12. Consider a regular pyramid T.ABCD having an edge-length of 6 cm. The
angle between plane TBC and base plane is 60o
. Point P is the midpoint of
TD. Plane ߙ is drawn passing through point B and P and parallel to AC.
Calculate tan ∠(ߙ ,base plane).
13. Given a pyramid D.ABC, where DB is perpendicular to the base plane,
∠ܤ = 90°, ܤܣ = 3 cm, ܥܤ = 4 cm, and ܣܦ = 5 cm. Calculate
(ܥܤܣ ,ܥܣܦ)∠ tan
14. Given a regular pyramid T.ABCD with lateral
edges TA = TC = TB = TD = 6 cm and base plane
edges AB = BC = CD = DA = 4 cm. If ߙ is the
angle between planes TAD and TBC, find the
value of sin ߙ .
15. Two equilateral triangles ABC and BCD are perpendicular to each other
and intersect each other at BC. If the side length of both triangles is 12 cm,
find the cosine of angle ABD.
Mathematics for Grade XII 49
Statistics
Statistics is a branch of
applied mathematics
which appeared and
flourished in the 19th
Century. Throughout its
development, statistica
has taken major roles in
various areas, from
science to practical
applications in industry
and sports.
Statistics has two major functions: presenting data in forms that are systematics,
easily understood, and easily described (called descriptive statistics) and as a tool
for drawning conclusions (called inference statistic).
7. Barangsiapa yang mengerjakan kebaikan seberat dzarrahpun, niscaya dia akan
melihat (balasan)nya. 8. Dan barangsiapa yang mengerjakan kejahatan sebesar
dzarrahpun, niscaya dia akan melihat (balasan)nya pula. (Q.S. Az Zalzalah ayat 7-8)
Descriptive Statistics
What is the purpose of lesson?
3.2 Mendeskripsikan dan menggunakan berbagai ukuran pemusatan, letak dan
penyebaran data sesuai dengan karakteristik data melalui aturan dan rumus serta
menafsirkan dan mengomunikasikannya.
4.2 Menyajikan dan mengolah data statistik deskriptif kedalam tabel distribusi dan
histogram untuk memperjelas dan menyelesaikan masalah yang berkaitan
dengan kehidupan nyata.
Chapter 2
Mathematics for Grade XII 50
A. Measuring the Center of Data
1.The Mean
A measure of location which does make use of the actual values of all the
observations is the mean. This is the quantity which most people are referring
to when they talk about the ‘average’. The mean is found by adding all the
values and dividing by the number of values.
Suppose that you wanted to know the typical playing time for a compact disc
(CD). You could start by taking a few CDs and finding out the playing time
for each one. You might obtain s list of values such as
49,56,55,68,61,52,63
mean =
49 + 56 + 55 + 68 + 61 + 52 + 63
7
The mean ݔ ,̅of a data set of ݊ values is given by
Calculating the mean from a frequency table
You can include this calculation in the table by adding a third column in
which each value of the variable, ݔ
is multiplied by its frequency ݂
.
The mean, ݔ ,̅of a data set in which the variable takes the value ݔଵ with
frequency ݂ଵ, ݔଶ with frequency ݂ଶ and so on is given by
Example 1
The following table gives the number of brothers and sisters of the children
at a school.
Solution
=̅ݔ
ݔ + ⋯ + ଶݔ + ଵݔ
݊
=
ݔ ∑
݊
=̅ݔ
ݔଵ݂ଵ + ݔଶ݂ଶ + ⋯ + ݔ݂
݂ଵ + ݂ଶ + ⋯ + ݂
=
݂ݔ ∑
∑ ݂
Mathematics for Grade XII 51
Number of brother
and sisters, ࢞
ࢌ࢞ ࢌ ,Frequency
0
1
2
3
4
5
6
36
94
48
15
7
3
1
0
94
96
45
28
15
6
Totals: ∑ ݂ = 204 ∑ ݔ = ݂284
The mean is equal to ଶ଼ସ
ଶସ
= 1.39
If the data in a frequency table are grouped, you need a single value to
represent each class before you can calculate the mean using equation
above. A reasonable choice is to take the value halfway between the class
boundaries. This is called the mid-class value.
Example 2
The following table show the playing times of 95 CDs. Two other columns
have been included, one giving the mid-class value for each class and the
other the product of this mid-class value and the frequency.
Solution
Playing Time,
ݔ) min) Class boundaries Frequency,
݂
Mid-class
ݔ ,value
݂ݔ
40 − 44
45 − 49
50 − 54
55 − 59
60 − 64
65 − 69
70 − 74
75 − 79
39.5 ≤ ݔ ≥ 44.5
44.5 ≤ ݔ ≥ 49.5
49.5 ≤ ݔ ≥ 54.5
54.5 ≤ ݔ ≥ 59.5
59.5 ≤ ݔ ≥ 64.5
64.5 ≤ ݔ ≥ 69.5
69.5 ≤ ݔ ≥ 74.5
74.5 ≤ ݔ ≥ 79.5
1
7
12
24
29
14
5
3
42
47
52
57
62
67
72
77
42
329
624
1368
1798
938
360
231
Totals:
݂ = 95
= ݂ݔ ∑
5690
Thus the estimate of the mean is ∑ ௫
∑
=
ହଽ
ଽହ
= 59.9 minutes.
Mathematics for Grade XII 52
1) The test marks of 8 students were 18, 2, 5, 0, 17, 15, 16, and 11. Find the
mean test mark.
2) The following table gives the frequency distribution for the lengths of
rallies (measured by the number of shots) in a tennis match.
Length of rally 1 2 3 4 5 6 7 8
Frequency 2 20 15 12 10 5 3 1
Find the mean length of a rally.
3) Calls made by a telephone saleswoman were obtained. The lengths (in
minutes, to the nearest minute) of 30 calls are summarized in the following
table.
Length of call 0 – 2 3 – 5 6 – 8 9 – 11 12 – 15
Number of calls 17 6 4 2 1
a) Write down the class boundaries.
b) Estimate the mean length of the calls.
4) The volumes of the contents of 48 half-litre bottles of orangeade were
measured, correct to the nearest milliliter. The results are summarized in
the following table.
Volume (ml) 480 – 489 490 – 499 500 – 509 510 – 519 520 – 529 530 – 539
Frequency 8 11 15 8 4 2
Estimate the mean volume of the contents of the 48 bottles.
2. The Median
Suppose that you wanted to know the typical playing time for a compact
disc (CD). You should start by taking a few CDs and finding out the
playing time for each one. You might obtain a list of values such as
49, 56, 55, 68, 61, 57, 61, 52, 63
Where the values have been given in minutes, to the nearest whole minute.
You can get a clearer picture of the location of the playing times by
arranging them in ascending order of size :
49, 52, 55, 56, 57, 61, 61, 63, 68
A simple measure of location is the ‘middle’ value, the value that has equal
numbers of values above and below it. In this case there are nine values
and the middle one is 57. This value is called the median.
To find the median of a data set of ݊ values, arrange the values in order of
increasing size. If ݊ is odd, the median is the ଵ
ଶ
(݊ + 1)th value. If ݊ is
Classroom Activities 1
Mathematics for Grade XII 53
even, the median is halfway between the ଵ
ଶ
݊th value and the following
value.
Data set are often much large that the ones in the previous section and the
values will often have been organized in some way, maybe in a frequency
table. As an example, the following table gives the number of brothers and
sisters of the children at a school.
Number of brother and
sisters
Frequency Cumulative Frequency
0
1
2
3
4
5
6
36
94
48
15
7
3
1
36
130
178
193
200
203
204
Totals: 204
The median of this data set is … .
The median of a data group frequency distribution table is determined by
using interpolation. Recall that in a group frequency distribution table we
lose the details of the original data such that we can no longer determine
the median precisely.
Observe the math test results in the following table.
Value Frequency Less-than Cumulative
Frequency
30 – 39
40 – 49
50 – 59
60 – 69
70 – 79
80 – 89
90 – 99
2
3
9
12
6
6
2
2
5
14
26
32
38
40
The size of data in table above is 40 (even), so the median lies between the
20th and the 21th-entries.
The median of a grouped data is determined by the interpolating formula
as follows:
Mathematics for Grade XII 54
I wanna try !
Where:
ܾ = lower boundary of the median class
݊ = data size
݂ = less-than cumulative frequency preceding the median class
݂ = frequency of the median class
݇ = class interval
For the data, we can obtain = 59.5, ݊ = 40, ݂ = 14, ݂ = 12, ݇ = 10,
hence,
Median = 59.5 + ቌ
1
2
∙ 40 − 14
12 ቍ ∙ 10 = 64.50
Determine the median of the set of data shown in the following frequency
distribution table.
Weight (kg) Frequency Less-than cumulative
frequency
40 – 49
50 – 59
60 – 69
70 – 79
80 – 89
5
14
16
12
3
…
…
…
…
…
3. The Mode
A third measure of location is the mode, sometimes called the modal
value. This is defined to be the most frequently occurring value. You can
pick it out from a frequency table (if the data have not been grouped) by
looking for the value with the highest frequency. If you are given a small
data set, then you can find the mode just by looking at the data. For the
first nine CDs in last section, with playing times
49, 52, 55, 56, 57, 61, 61, 63, 68
The mode is 61.
It is not uncommon for all the values to occur only once, so that there is
no mode. For example, the next six CDs had playing times
47, 49, 59, 62, 65, 68
Median = ܾ + ቌ
1
2
݊ − ݂
݂
ቍ ∙ ݇
Mathematics for Grade XII 55
And there is no modal value. Combining the two data sets gives
47, 49, 49, 52, 55, 56, 57, 59, 61, 61, 62, 63, 65, 68, 68.
Now there are three values which have a frequency of 2, giving three
modes: 49, 61, and 68.
If data have been grouped, then it only possible to estimate the mode.
Alternatively, you can give the modal class, which is the class with the
highest frequency density.
Consider the following table.
Test Score Midpoint (࢞ (Frequency (ࢌ(
55 – 59
60 – 64
65 – 69
70 – 74
75 – 79
80 – 84
85 – 89
90 – 94
57
62
67
72
77
82
87
92
7
12
23
21
18
10
8
1
The modal class of these data is the class 65 – 69. A more certain method
to determine the mode is through interpolation. The mode value can be
determined by the following formula.
Where:
ܾ = lower boundary of the median class
݀ଵ = the difference in the frequencies of the modal class and the preceding
class
݀ଶ = the difference in the frequencies of the modal class and the following
class
݇ = class interval
For the data, we can obtain ܾ = 64.5, ݀ଵ = 23 − 12 = 11 , ݀ଶ = 23 −
21 = 2 , ݇ = 5, hence
Mode = 64.5 + ൬
11
11 + 2൰ ∙ 5 = 68,73
Mode = ܾ + ൬
݀ଵ
݀ଵ + ݀ଶ
൰ ∙ ݇
Mathematics for Grade XII 56
I wanna try !
Determine the mode of the data presented in the following frequency
distribution table.
a. Value Frequency b. Value Frequency
31 – 40
41 – 50
51 – 60
61 – 70
71 – 80
81 – 90
91 – 100
4
7
11
16
13
9
1
36 – 42
43 – 49
50 – 56
57 – 63
64 – 70
71 – 77
78 – 84
85 – 91
9
13
15
21
13
11
5
3
B. Measures of Position Data Clusters
1. The Quartile
Quartile divide data into four equal parts. There are three quartiles, namely
the lower quartile (ܳଵ), the middle quartile (ܳଶ) or median and the upper
quartile (ܳଷ). Recall that quartiles can be determined if the data are already
ordered.
Example 4
Determine ܳଵ, ܳଶ, and ܳଷ from sets of data below:
7 8 22 20 15 18 19 13 11
Solution
First, arrange the data in numerical order
7 8 11 13 15 18 19 20 22
Since the number of data values (9) is odd, find the median ܳଶ = 15 and
delete it.
7 8 11 13 18 19 20 22
This automatically divides the data into lower and upper halves.
The median of the lower half is the lower quartile, so ܳଵ =
ଵ
ଶ
(8 + 11) =
9.5, and the median of the upper half is the upper quartile, so ܳଷ =
ଵ
ଶ
(19 + 20) = 19.5.
The interquartile range is ܳଷ − ܳଵ = 19.5 − 9.5 = 10
We have discussed how to determine ܳଵ,ܳଶ, and ܳଷ for a single data. In
order to estimate the quartile of a set of grouped data, we will use
interpolation like in the case of determining the median.
Mathematics for Grade XII 57
ܳ = ܾ + ቆ
భ
ర
ିೖೖೞ
ೂ
ቇ ∙ ݇
Where:
ܾ = lower boundary of the ݅-th quartile
݊ = data size
݂௦ = less-than cumulative frequency before the ݅-th quartile class
݂ொ
= frequency of the ݅-th class
݇ = class interval
Example 5
Given the following data:
Weight (kg) Frequency
40 – 49 5
50 – 59 14
60 – 69 16
70 – 79 12
80 – 89 3
Total 50
Determine ܳଵ,ܳଶ, and ܳଷ from the data above.
Solution
The table above is completed with the following cumulative frequencies.
Data size (݊) = 50
Weight (kg) Frequency Cumulative
frequency
40 – 49 5 5
50 – 59 14 19 } ⟶ ܳଵ
60 – 69 16 35 } ⟶ ܳଶ
70 – 79 12 47 } ⟶ ܳଷ
80 – 89 3 50
Total 50
ଵ
ସ
∙ (50) = 12.5 ⟹ ܳଵ, corresponding to class 50 – 59
ଵ
ଶ
∙ (50) = 25 ⟹ ܳଶ, corresponding to class 60 – 69
ଷ
ସ
∙ (50) = 37.5 ⟹ ܳଷ, corresponding to class 70 – 79
Mathematics for Grade XII 58
For class ܳଵ(50 − 59), ܾଵ = 49.5; ݂௦భ = 5; ݂ொభ = 14; ݇ = 10
hence ܳଵ = ܾଵ + ቆ
భ
ర
ିೖೖೞభ
ிೂ
ቇ ∙ ݇ = 49.5 + ቀ
ଵଶ.ହିହ
ଵସ ቁ ∙ 10
= 54.86
For class ܳଵ(50 − 59), ܾଵ = 49.5; ݂௦భ = 5; ܨொభ = 14; ݇ = 10
hence ܳଵ = ܾଵ + ቆ
భ
ర
ିೖೖೞభ
ிೂభ
ቇ ∙ ݇ = 49.5 + ቀ
ଵଶ.ହିହ
ଵସ ቁ ∙ 10
= 54.86
For class ܳଶ(60 − 69), ܾଵ = 59.5; ݂௦మ = 19; ܨொమ = 16; ݇ = 10
hence ܳଶ = ܾଵ + ቆ
భ
మ
ିೖೖೞమ
ிೂమ
ቇ ∙ ݇ = 59.5 + ቀ
ଶହିଵଽ
ଵ ቁ ∙ 10
= 63.25
For class ܳଷ(70 − 79), ܾଵ = 69.5; ݂௦య = 35; ܨொయ = 12; ݇ = 10
hence ܳଷ = ܾଵ + ቆ
య
ర
ିೖೖೞయ
ிೂయ
ቇ ∙ ݇ = 69.5 + ቀ
ଷ.ହିଷହ
ଵଶ ቁ ∙ 10
= 71.58
2. Decile and the Percentile
Two other measures of data that are also often used are deciles
(ܦଵ,ܦଶ, … ܦଽ) and percentiles (ܲଵ, ܲଶ, … ܲଽଽ) which respectively divide
data into 10 and 100 parts. The procedure of determining deciles and
percentiles is the same as the one used in determining quartiles, except that
the ቀ
ଵ
ଶ
ቁ ݊ is replaced with ቀ
ଵ
ଵቁ ݊ or ቀ
ଵ
ଵቁ ݊.
Deciles of grouped data are determined by the following formula.
ቌ = ܾ + ܦ
݉
10 ݊ − ݂௦ௗ
݂ ቍ ∙ ݇
Where: ݉ = 1, 2, 3, … 9
ܾ = lower boundary of the ݉௧ −decile class
݊ = data size
݂௦ௗ =less-than cumulative frequency before the ݉௧ −decile class
݂ = frequency from the ݉௧ −decile class
݇ = class interval
While percentiles are determined by the following formula.
ܲ = ܾ + ቌ
݉
100 ݊ − ݂௦
݂
ቍ ∙ ݇
Mathematics for Grade XII 59
Where: ݉ = 1, 2, 3, … 99
ܾ = lower boundary of the ݉௧ −percentile class
݊ = data size
݂௦ =less-than cumulative frequency before the ݉௧ −percentile class
݂ = frequency from the ݉௧ −percentile class
݇ = class interval
Example 6
Given the following data.
Weight (kg) Frequency
40 – 49 5
50 – 59 14
60 – 69 16
70 – 79 12
80 – 89 3
Total 50
Determine ܦଽ
, ܲଷ, and ܲଽଽ.
Solution
ܦଽ = ܾଽ + ቆ
వ
భబ
ିೖೖೞ
ವవ ቇ ∙ ݇
= 69.5 + ቀ
ସହିଷହ
ଵଶ ቁ ∙ 10
= 77.83
For ܲଷ
ଷ
ଵ
∙ 50 = 15, class of ܲଷ is 50 – 59, hence ܾଷ = 49.5; ݂௦ =
5; ݂యబ = 14; ݇ = 10, hence:
ܲଷ = ܾଷ + ൬
భబబ
ିೖೖೞ
ುయబ
൰ ∙ ݇
= 49.5 + ቀ
ଵହିହ
ଵସ ቁ ∙ 10
= 56.64
For ܲଽଽ
ଽଽ
ଵ
∙ 50 = 49.5, class of ܲଽଽ is 80 – 89, hence ܾଽଽ = 79.5; ݂௦ =
47; ݂వవ = 3; ݇ = 10, hence:
ܲଽଽ = ܾଽଽ + ቆ
వవ
భబబ
ିೖೖೞ
ುవవ ቇ ∙ ݇
= 79.5 + ቀ
ସଽ.ହିସ
ଷ
ቁ ∙ 10
= 87.83
Mathematics for Grade XII 60
1. Determine the median and quartiles of the data given in the following
distribution tables.
a. Value Frequency b. Value Frequency
41 – 45 12 36 – 42 9
46 – 50 15 43 – 49 13
51 – 55 17 50 – 56 15
56 – 60 23 57 – 63 21
61 – 65 13 64 – 70 13
66 – 70 10 71 – 77 11
71 – 75 6 78 – 84 5
76 – 80 4 85 – 91 3
2. Determine the median and quartiles of the data given by the following
histograms.
C. Measures of Spread
1. The Range
The range of a set of data values is defined by the equation
Range = largest value ± smallest value
I wanna try !
Mathematics for Grade XII 61
Example
Consider the two sets of data given below:
48 52 60 60 68 72
The range of data set is 72 – 48 = 24
2. The Interquartil Range
Interquartil range = upper quartile − lower quartile = ܳଷ − ܳଵ
Example 7
Find the quartiles and the interquartile range for each of the sets of data
below
7 8 22 20 15 18 19 13 11
Solution
In the last example, we obtain ܳଵ = 9.5, ܳଷ = 19.5
The interquartile range is ܳଷ − ܳଵ = 19.5 − 9.5 = 10
3. Variance and Standard Deviation
Measures of data distribution that are most commonly used in statistics are
variance and standard deviation. Variance and standard deviation give
descriptions of data variation around the mean value. Because the mean is
the value that represents the whole data and is the main focus, we should
expect that some observed values will be smaller than the mean and some
will be larger.
Consider a set of data which contain the following values.
1 3 8 10 13
The mean, x, of the data given above is 7 and the deviations of each of its
entries x x i are
−6 −4 1 3 6
Note that the sum of the deviations is always zero.
In order too avoid misinterpretation regarding the measure of data
distribution given by (unit)ଶ
, we use a measure called standard deviation.
Standard deviation measures data ution in the same units as the units on
the data.
Variance S
2 in a set of data ݔଵ, ݔଶ, … , ݔ is the mean of the sum of the
square of the deviatins of each entry, or
n
i
S xi x n 1
2
2 1
Standard deviation is the quare-root of variance, or
n
i
xi x n
S
1
1 2
Mathematics for Grade XII 62
The formulas for variance and standard deviation given above are used to
determine the variance and standard deviation in a ser od data obtained from
a population. However, sometimes the data obtained from a population are
too large. In such cases we only use samples taken from among the data.
Finding the variance and standard deviation from sample data is as follows.
Example 8
Determine the variance and standard deviation in the following grouped
data certain populations.
a. 2 3 6 8 11
b. 11 12 13 14 15 16 17 18
Solution
a. 6
5
30
x
ݔ ݔ x (ݔ x )
2
2 2 – 6 = -4 16
3 3 – 6 = -3 9
6 6 – 6 = 0 0
8 8 – 6 = 2 4
11 11 – 6 =5 25
ݔ = 30 (ݔ x )
ଶ = 54
ܵ
ଶ =
ଵ
ହ
(̅ݔ − ݔ) ∑
ଶ =
ହସ
ହ
= 10.8
ୀଵ
ܵ = √10.8 ≈ 3.3 (one decimal places)
Hence, the variance is 10.8 and the standard deviation 3.3.
Variance S
2 of samples is
n
i
S xi x n 1
2
2
1
1
Standard deviation (S) of samples is
n
i
xi x n
S
1
2
1
1
Mathematics for Grade XII 63
b. 14.5
8
116
x
ݔ ݔ x (ݔ x )
2
11 11 – 14.5 = -3.5 12.25
12 12 – 14.5 = -2.5 6.25
13 13 – 14.5 = -1.5 2.25
14 14 – 14.5 = -0.5 0.25
15 15 – 14.5 = 0.5 0.25
16 16 – 14.5 = 1.5 2.25
17 17 – 14.5 = 2.5 6.25
18 18 – 14.5 = 3.5 12.25
ݔ = 116 (ݔ x )
ଶ = 42
ܵ
ଶ =
∑(௫ x )
మ
=
ସଶ
଼
= 5.25
ܵ = √5.25 ≈ 2.3 (one decimal places)
Hence, the variance is 5.25 and the standard deviation 2.3.
A different formula that can be used to obtain variance is
ܵ) ܍܋ܖ܉ܑܚ܉܄
ଶ) =
1
ݔ݊
̅ݔ − ଶ
ଶ
ୀଵ
Example 9
Try to solve the following set of data by using the following method
11 12 13 14 15 16 17 18
Solution
ݔ ݔ
ଶ
x=
∑ ௫
=
ଵଵ
଼
= 14.5
ܵ
ଶ =
1
ݔ݊
̅ݔ − ଶ
ଶ
ୀଵ
=
1
8
(1.742) − (14.5)ଶ
= 215.5 − 210.25
= 5.25
ܵ = √5.25 ≈ 2.3
11 121
12 144
13 169
14 196
15 225
16 256
17 289
18 324
ݔ 116 = ݔ
ଶ = 1,724
Mathematics for Grade XII 64
The variance of grouped data is also represented by the formula given
above where ݔ
represents the midpoint value of the ݅
௧-class.
Example 10
Determine the variance and standard deviation in the following
population’s data.
Value Frequency
141 – 147 2
148 – 154 7
155 – 161 12
162 – 168 10
169 – 175 9
176 – 182 7
183 - 189 3
Solution
Method 1
Value Midpoint
(ݔ)
Frequency
(݂)
݂ݔ ݔ x (ݔ x
)
2
݂(ݔ x )
2
141 – 147 144 2 288 –21 41 882
148 – 154 151 7 1,057 –14 196 1,372
155 – 161 158 12 1,896 –7 49 588
162 – 168 165 10 1,650 0 0 0
169 – 175 172 9 1,548 7 49 441
176 – 182 179 7 1,253 14 196 1,372
183 – 189 186 3 558 21 441 1,323
ݔ݂ 50 ݂ =
= 8,250
݂ ቀݔ x ቁ
ଶ
= 5,978
Mean x =
∑ ௫
∑
=
଼,ଶହ
ହ
= 165
Variance (ܵ
ଶ) =
ଵ
ݔ ∑
̅ݔ − ଶ
ଶ
ୀଵ
=
ହ,ଽ଼
ହ
= 119,56
Standard deviation (ܵ) = ඥ119,56 ≈ 10,9
Mathematics for Grade XII 65
Method 2
Value Frequency
(݂)
Midpoint
(ݔ)
ݔ݂ ݔ݂
2
141 – 147 2 144 288 41,472
148 – 154 7 151 1,057 159,607
155 – 161 12 158 1,896 299,565
162 – 168 10 165 1,650 272,250
169 – 175 9 172 1,548 266,256
176 – 182 7 179 1,253 224,287
183 – 189 3 186 558 108,788
ݔ݂ 50݂ =
= 8,250
ݔ݂
ଶ
= 1,367,228
Mean x =
∑ ௫
∑
=
଼,ଶହ
ହ
= 165
Variance (ܵ
ଶ) =
ଵ
ݔ݂ ∑
̅ݔ − ଶ
ଶ
ୀଵ
=
ଵ
ହ
(1,367,228) − (165)ଶ = 119,56
Standard deviation (ܵ) = ඥ119,56 ≈ 10,9
Hence, the variance in the data is 119.56 and the standard deviation is 1.09.
1. Determine the variance and standard deviation in the following
population’s data
39 43 44 45 46 47
2. Determine the variance and standard deviation in the following
population’s data
Value Frequency
3
4
5
6
7
4
13
13
7
3
Classroom Activities 2
Mathematics for Grade XII 66
3. Determine the variance and standard deviation in the following sample
data.
Value Midpoint (ݔ (Frequency (݂)
21 – 15
26 – 30
31 – 35
36 – 40
41 – 45
46 – 50
51 – 55
56 – 60
61 – 65
23
28
33
38
43
48
53
58
63
7
15
21
19
17
9
5
3
2
Classroom Activities 2
Mathematics for Grade XII 67
CHAPTER REVIEW EXERCISE
1. The arithmetic mean of math test results of 40 students is 8,2. If another student is
added, the aritmethic mean becomes 8,24. The value of the added student is … .
A. 10 D. 8,7
B. 9,8 E. 8,5
C. 9,6
2. The following tabel lists the results of a certain test.
Value Frequency
30 4
40 6
50 3
60 12
70 15
80 8
90 2
The number of students who have the value more than arithmetic mean is … .
A. 10 D. 37
B. 13 E. 40
C. 25
Look at the following frequency distribution table to solve questions number 3
and 4
Value Frequency
31 – 36 4
37 – 42 6
43 – 48 9
49 – 54 14
55 – 60 10
61 – 66 5
67 – 72 2
3. The median in the data is … .
A. 54,50 D. 51,02
B. 51,50 E. 50,07
C. 51,07
4. The mode of the data is … .
A. 49,06 D. 51,33
B. 50,20 E. 51,83
C. 50,70
Mathematics for Grade XII 68
5. The histogram below shows weight data of 50 students. The mean of the data is … .
Look at the following frequency distribution table to solve questions number 6
and 7
Value Frequency
30 – 39 2
40 – 49 4
50 – 59 5
60 – 69 9
70 – 79 10
80 – 89 7
90 – 99 3
6. The lower quartile and the upper quartil of data is … .
A. 56,5 and 79,5 D. 59,5 and 78,5
B. 57,5 and 78,5 E. 60,5 and 79,5
C. 58,5 and 79,5
7. The semi-interquartil of data is … .
A. 23 D. 11,5
B. 21 E. 10,5
C. 13,5
8. Given data : 7, 4, 7, 5, 3, 9, 8, 7, 6, 4.
(1) Range = 6
(2) Mean = 6
(3) Mode = 7
(4) Median = 6,5
The true statements from above data is … .
A. (1), (2), and (3) D. only (4)
B. (1) and (3) E. all
C. (2) and (4)
0
2
4
6
8
10
12
14
16
42 47 52 57 62 67
Weight (kg) Frequency
A. 52,5
B. 55,5
C. 55,8
D. 60,3
E. 60,5
Mathematics for Grade XII 69
9. Standart deviation of the following frequency distribution table is … .
Weight
(kg)
Frequency
(݂)
Midpoint
(ݔ)
൫ݔ − x൯ ൫ݔ − x൯
ଶ
݂൫ݔ − x൯
ଶ
43 – 47 5 45 −7 49 245
48 – 52 12 50 −2 4 48
53 – 57 9 55 3 9 27
58 – 62 4 60 8 64 384
A. 23,4 D. √23,4
B. 26,5 E. ඥ26,5
C. 21
10. A botanist collects data of new rose varieties. For every sample plant, he recorded
the number of days taken for the flowers to bloom. The result are shown in the
following table.
Value Frequency
15 – 19 14
20 – 24 16
25 – 29 12
30 – 39 10
40 – 49 8
The value of ܲଷହ is … .
A. 20,67 D. 25,33
B. 21,15 E. 30,44
C. 21,69
11. Look at the positive-ogive below and then determine the mean, median, and mode.
0
5
10
15
20
25
30
35
40
45
50
55
0 10 20 30 40 50 60 less-than cumulative frequency
Upper boundaries
Possitive-ogive
Mathematics for Grade XII 70
12. Look at the histogram below and then determine the interquartile range, ܦସ, and
଼଼ܲ.
13. Fill the table below and then calculate the variance and standart deviation.
(̅ݔ − ݔ) ̅ݔ − ݔ ݔ݂ ݔ ݂ Value
(̅ݔ − ݔ)݂ ଶ
ଶ
1-3 1
4-6 2
7-9 3
10-12 4
13-15 5
0
7
14
21
28
35
42 FREQUENCY
MID-POINT
Histogram
Mathematics for Grade XII 71
Probability i
We often face the problems
associated with order,
arrangement or the like. In such
cases there are two types of
arrangement that is arrangement
with respect to the other and
arrangement which does not pay
attention to the order.
In the case of mathematical structure which respect of the order called a permutation,
while those not reffered of the other called combination. This chapter will explain
about permutation and combination.
Sesungguhnya Kami menciptakan segala sesuatu menurut ukuran. (Q.S. Al Qamar
ayat 49)
What is the purpose of lesson?
3.3 Menganalisis aturan pencacahan (aturan penjumlahan, aturan
perkalian, permutasi, dan kombinasi) melalui masalah kontekstual
4.3 Menyelesaikan masalah kontekstual yang berkaitan dengan kaidah
pencacahan (aturan penjumlahan, aturan perkalian, permutasi, dan
kombinasi)
Chapter 3
Mathematics for Grade XII 72
Counting rules are rules used to count all probabilities that may occur
in a certain experiment. We will study three counting methods, namely the
filling possibilities rules, the permutation rules, and the combination
method. Before we continous, observe the following events that relate to
everyday life.
(1) Two candidates are to be selected as the president and vice-president of
the student council from among top four candidates in a school. An
honorary council is formed to carry out the selection. The honorary council
consists of representatives from each class. How many combinations of
the president and vice president that the council must consider? Suppose
the candidates are Daffa, Rafly, Reza, and Erwin.
(2) Rafif plans on vacation to Bali. He considers taking three novels from six
novels that he just bought. He cannot make up his mind on which three
novels he should take. How many combinations of three novels that can
be considered?
The two events above are just examples of problems that can be solved by
using counting rules.
A. Filling Possibility Rules
In problem-solving that involves the filling possibility rules, we
manually list all probabilities. There are several ways of listing in this rule,
three of them will be discussed here, namely tree diagram, cross tables, and
ordered pairs.
1. Tree Diagrams
We will solve problem (1) above by using tree diagram. We notice
that in selecting the pairs, the pair (Daffa, Rafly) is to be distinguished
from the pair (Rafly, Daffa) because the name that is mentioned first
becomes the president, whereas the name that is mentioned second
becomes the vice-president. Below is the list of all possibilities to select
the president and the vice-president.
Mathematics for Grade XII 73
President Vice-president Pair
(Daffa, Rafly)
(Daffa, Erwin)
(Daffa, Reza)
(Rafly, Daffa)
(Rafly, Erwin)
(Rafly, Reza)
(Erwin, Daffa)
(Erwin, Rafly)
(Erwin, Reza)
(Reza, Daffa)
(Reza, Rafly)
(Reza, Erwin)
From the tree diagrams above, the honorary council must consider
12 pairs of president and vice-president combinations.
Problem (2). Suppose the novels are A, B, C, D, E, and F. Three
novels are to be chosen from among these six. Notice that the choice ( A
B C) is the same as (B C A), (A C B), (B A C), (C A B), or (C B A).
Determine the possibilities combinations of three novels that can be
considered?
2. Cross Tables
Steps in constucting cross table. Suppose we want to solve problem (1) by
using cross tables. Write the first component (candidate forpresident) in
columns and the second component in rows. The (column, row) pairs show
allposible outcomes of the selection process.
Mathematics for Grade XII 74
Vice President
President
Daffa Rafly Erwin
Reza
Daffa - (Daffa,
Rafly)
(Daffa,
Erwin)
(Daffa,
Reza)
Rafly (Rafly,
Daffa)
- (Rafly,
Erwin)
(Rafly,
Reza)
Erwin (Erwin,
Daffa)
(Erwin,
Rafly)
- (Erwin,
Reza)
Reza (Reza,
Daffa)
(Reza,
Rafly)
(Reza,
Erwin)
-
By considering all possible pairs, we can conclude the honorary council
must consider 12 pairs of (president, vice-president). Cross tables seem
more difficult to use in cases that involve a lot more probabilities, such as
selecting 11 out of 22 football players. Moreover, cross tables can only be
used to select pairs. Therefore, problem (2) cannot be solved by using this
method.
3. Ordered Pairs
The method of ordered pairs is the easiest and least tedious to use.
Problem (1) can be solved through ordered pairs as follows. Suppose ܣ=
{Daffa, Rafly, Erwin, Reza} is a set of candidates for president and vicepresident. By insisting that no one is allowed to filltwo positions that the
pair (ݔ ,ݕ (is different from (ݕ ,ݔ ,(the ordered pairs of ܣ are: {(Daffa,
Rafly), (Daffa, Erwin), (Daffa, Reza), (Rafly, Daffa), (Rafly, Erwin),
(Rafly, Reza), (Erwin, Daffa), (Erwin, Rafly), (Erwin, Reza), (Reza,
Daffa), (Reza, Rafly), (Reza, Erwin)}. The number of ordered pairs from
ܣ is 12. Hence, the honorary council must consider 12 combinations for
(president, vice-president)
Problem (2): Suppose ܰ = {ܣ ,ܤ ,ܥ,ܦ ,ܧ ,ܨ {represents six novels from
which three are to be chosen by Andi. Because the selection does not
Mathematics for Grade XII 75
distinguish the order, meaning that (ܥܤܣ) = (ܥܣܤ (etc., the combinations
that can be formed are: (Write in the space beside)
So, the total number of possibilities that can occur is ....
In solving problem (1) above, it appears as if we can carry out the selection
in two stages, namely:
1. When selecting the president out of four candidates.
2. When selecting the vice-president from th eremaing unchosen three
candidates. Because each of the four candidates may be paired-off with the
remaining three, the number of pairs 4 × 3 = 12 ways.
4. Summation Rules and Multiplication Rules
a. Summation Rules
Suppose an event can occur in ݊ different ways (mutually exclusive).
The first way has ଵ different possibilities, the second way has ଶ
different possibilities and so on until the ݊
௧-way which has different
possibilities. Hence the total number of possibilities in the event is
. + ⋯ + ଶ + ଵ
b. Multiplication Rules
If an event consists of ݊ ordered stages where the first stage can occur
in ݍଵ different ways, the second stage can occur ݍଶ different ways, the
total number of ways that can occur in the event is ݍଵ × ݍଶ × … × ݍ .
Example 1
City ܣ and city ܤ are connected by three different alternative routes. City ܤ
and city ܥ are also connected with three different routes. When we travel from
ܣ to ܥ through ܤ ,how many different routes can we take?
Mathematics for Grade XII 76
Solution
The trip from ܣ to ܥ through ܤ is
completed in two stages.
The first stage is when going from ܣ
to ܤ which can be done in 3 ways.
The second stage is when going from
ܤ to ܥ which can also be done in 3
ways.
According to the multiplication rules,
the total numbe of routes in going
from ܣ to ܥ is 3 × 3 = 9.
Example 2
A number is to be constructed from the numbers 1, 2, 3, 4, and 5.
a. How many different three-digit numbers can be constructed?
b. How many numbers less than 400 (no repeated numbers) can be
constructed?
c. How many three-digit numbers more than 430 can be constructed?
Solution
a. Three-digit number contain hundreds (H), tens (T), and single units (U).
So we arrange the possibilities in the same order.
Digit value H T U
Numbers of ways 5 4 3
According to the multiplication ruless, the total number of different threedigit numbers that can be constructed is 5 × 4 × 3 = 60.
b. Numbers that are less than 400 may contain one, two, or three digits.
There are 5 single-digit numbers that can be constructed.
Digit value U
Numbers of ways 5
Mathematics for Grade XII 77
There are 5 × 4 = 20 two-digit numbers that can be constructed.
Digit value T U
Numbers of ways 5 4
Three digit numberscan be constructed in 3 × 4 × 3 = 36 ways.
Digit value H T U
Numbers of ways 3 4 3
According to the summation rules, the total non-repeatable numbers
less than 400 that can be constructed are 5 + 20 + 36 = 61.
c. There are many ways to determine the total non-repeatable numbers less
than 430. We will shown the simple way.
Numbers more than 400 can be constructed in 2 × 4 × 3 = 24 ways.
Digit value H T U
Numbers of ways 2 4 3
Numbers between 400 and 430 can be constructed in 1 × 2 × 3 = 6
ways.
Digit value H T U
Numbers of ways 1 2 3
The total non- repeatable numbers less than 430 are 24 − 6 = 18.
Example 3
The number of hundreds of digits with the first digit and the second digit having
a difference of 2 is ... .
A. 120 C. 140 E. 160
B. 130 D. 150
Solution
Hundreds of digits consist of 3 digits.
The hundreds and tens digits difference of 2 is:
(1,3),(3,1), (2,4), (4,2), (3,5), (5,3), (4,6),(6,4), (5,7), (7,5),(6,8), (8,6), (7,9),
(9,7), (2,0)
Mathematics for Grade XII 78
Hundred of digits Ten of digits unit
15 10
The number of numbers= 15 × 10 = 150 (D)
1. Construct tree diagram to show all possible ways to arrange shirts of size
S (small), M (medium), L (large), and XL (extra large) in three different
colors of blue, green, and grey.
2. There are 7 routes city A with city V, and there are 8 routes city B with
city C. How many ways can be taken from A to C through B?
3. A students has four shirts of different colors: white, brown, blue, and
yellow. He also has three pairs of trousers of different colors: brown, blue,
and black. If the student has two pairs of shoes in black and brown colors,
in how many ways can the students combine the shirt, trousers, and shoes
of different colors?
4. Three-digit numbers are to be constructed from the numbers: 0, 1, 2, 3, and
4. How many numbers can be constructed if:
a. the numbers may be repeatable
b. the numbers may not be repeatable
c. the numbers may not be repeatable and the values shall be more than
230
5. A class of 40 students is to choose their class officials which consist of a
leader, a secretary and a treasurer. In how many ways can the class officials
be chosen if all students have equal rights to be chosen.
Classroom Activities
1
Mathematics for Grade XII 79
B. Factorial notation
Factorial notation can be understood throughh the following definition
Definition
1. Suppose that is a real number, then
∙ ∙ ... ( − )( − ) = !
2. ! =
From the definition above, by using the assocative characteristics of
multiplication, we can obtain the following relation:
݊! = ݊(݊ − 1)(݊ − 2) … 3 ∙ 2 ∙ 1
݊! = ݊(݊ − 1)!
݊! = ݊(݊ − 1)(݊ − 2)! , etc
Example 1
Determine the following values.
a. 6! b. 3! + 5! c. 3! × 5! d. 3! × 5!
Solution
a. 6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720
b. 3! + 5! = 3 ∙ 2 ∙ 1 + 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 6 + 120 = 126
c. 3! × 5! = 6 × 120 = 720
d. ଵ!
ଷ!!
=
ଵ∙ଽ∙଼
ଷ∙ଶ∙ଵ
= 120
How about ሾ(−6)!ሿ?
Example 2
Express each of the following in factorial notation.
a. 7 × 6 × 5 b. ଽ∙଼∙∙
ଵ∙ଶ∙ଷ
c. ݊(݊ − 1) … (݊ − ݇ + 1)
Solution
a. 7 ∙ 6 ∙ 5 = 7 ∙ 6 ∙ 5 ∙ ସ!
ସ!
=
!
ସ!
b. ଽ∙଼∙∙
ଵ∙ଶ∙ଷ
=
ଽ∙଼∙∙∙ହ!
ଷ!ହ!
=
ଽ!
ଷ!ହ!
c. ݊(݊ − 1) … (݊ − ݇ + 1) =
(ିଵ)…(ିାଵ)∙(ି)!
(ି)!
=
!
(ି)!
Mathematics for Grade XII 80
Example 3
Ari and Ira are members of a group of 9 people. The number of ways to
make a sequence provided that Ari and Ira are not side by side is ... .
A. 7 × 8! C. 5 × 8! E. 6 × 7!
B. 6 × 8! D. 7 × 7!
Solution
Number of rows of 9 people = 9!
The number of rows with Ari and Ira side by side = 2 × 8!
Then the number of ways to make a sequence provided that Ari and Ira are
not side by side is 9! − (2 × 8!) = (9 × 8!) − (2 × 8!) = (9 − 2) × 8! =
7 × 8! (A)
1. Determine the values of the forms:
a.
!
ଷ!
b. ଼!
ସ!ହ!
c.
ଵହ!ଽ!
ଵଶ!ଵ!
2. Express in factorial notation:
a. 12 ∙ 11 ∙ 10 ∙ 9 ∙ 8 b. ∙଼∙ଽ
∙ହ
c.
(ିଵ)(ିଶ)
ଵ∙ଶ∙ଷ∙ସ
3. Simplify the following forms:
a.
!
(ିଶ)!
, for ݊ ≥ 2 b. (ିଶ)!
(ାଶ)!
, for ݇ ≥ 2 c.
(ିାଵ)!
(ି)!
, for ݊ ≥ ݇
4. Determine the value of ݊ which satisfies the following equations:
a.
(ିସ)!
!
=
ଵ
ଵଶ
b. (ାଶ)!
(ିସ)!
= 20 c.
(ାଵ)!
(ିଵ)!ଶ!
=
!
(ିଶ)!
Classroom Activities 2
Mathematics for Grade XII 81
C. Permutation
Definition
The permutation of a group of subjects is the number of arrangements
of different objects in a certain order without repeating any of the
objects. Suppose ࡴ is a set of objects, and ≥ ,the permutation of
objects from the set ࡴ is the different object arrangements in a certain
order which consists of member objects of ࡴ .
1. Permutation of objects from different objects
Situation : There are ݊ different objects
Problem : determining the number of ordered arrangements
consisting of the existing ݊ objects
Notation : ܲ or ܲ
ܲ
= ݊!
Example 5
Out of four candidates for the student council officials, how many possible
appointments can occur in fnding the president, vice-president, treasurer,
and secretary?
Solution
This problem is a problem of permutation of 4 objects out of 4 objects.
ܲସ
ସ = 4! = 4 ∙ 3 ∙ 2 ∙ 1 = 24 possibilities. Thus, there are 24 possible
combinations of student council officials.
Example 6
Determine how many letter arrangements can be made out of the word
“QOMAR” if the arrangemenets are to be out of five different letters (no
letter is used twice).
Solution
This problem is a problem of permutation of 5 objects out of 5 objects.
Mathematics for Grade XII 82
ܲହ
ହ = 5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120 possible letter arrangemnets.
Thus, there are 120 possible letter arrangements.
2. Permutation of objects from different objects, ≥
Situation : There are ݊ different objects
Problem : determining the number of ordered arrangements of ݇
objects from ݊ objects that are available
Notation : ܲ or ܲ
ܲ
=
݊!
(݊ − ݇)!
Example 7
Determining the number of possibilities in selecting a president and vicepresident out of 5 candidates.
Solution
This problem is a problem of permutation of 2 objects out of 5 objects.
ܲଶ
ହ =
ହ!
(ହିଶ)!
=
ହ!
ଷ!
= 5 × 4 = 20 possibilities.
3. Permutation of objects from different objects which share
identical objects
Situation : There are ݊ objects some of which are identical. Suppose
there are ݊ଵ of object ݍଵ, ݊ଶ of object ݍଶ, … , ݊ of object
ݍ ,where ݊ଵ + ݊ଶ + ⋯ + ݊ = ݊
Problem : determining the number of ordered arrangements of ݊
objects
Notation : ܲ(݊ଵ, ݊ଶ, … , ݊)
ܲ(݊ଵ, ݊ଶ, … , ݊
) =
݊!
݊ଵ! ݊ଶ! … ݊!
Example 8
How many different letter combinations can be constructedfrom the word
“ABELARDIRA”?
Mathematics for Grade XII 83
Solution
There are 10 letters in the word ABELARDIRA; consisting of 3 of the
letter A, 2 of the letter R, 1 of letter B, 1 of letter E, 1 of letter L, 1 of letter
D, and 1 of letter I.
The number of letter combinations that can be contructed is:
ଵܲ(ଷ,ଶ,ଵ,ଵ,ଵ,ଵ,ଵ) =
ଵ!
ଷ!ଶ!ଵ!ଵ!ଵ!ଵ!ଵ!
=
ଵ∙ଽ∙଼∙∙∙ହ∙ସ
ଶ∙ଵ
= 302.400
Example 9
A library staff is to arrange three identical mathematics books, two
identical economics books, and four identical English literature books in a
row on a shelf. How many possible different arrangements can be made?
Solution
The number of possible arrangements is
ଽܲ(ଷ,ଶ,ସ) =
ଽ!
ଷ!ଶ!ସ!
= 1.260 ways.
4. Circular Permutation
Situation : There are ݊ different objects
Problem : Determining the number of ways ݊ different objects can be ordered
in a circular fashion.
Notation : ܲ(ୡ୧୰ୡ୳୪ୟ୰)
ܲ(ୡ୧୰ୡ୳୪ୟ୰) = (݊ − 1)!
Example 10
Azza, Jihan, Widad, Bila, Lia, and Salsa are going to have a meeting at a
circular table. How many different ways are there for them to be seated?
Solution
This problem isa problem of circular permutation with ݊ = 6.
ܲ(ୡ୧୰ୡ୳୪ୟ୰) = (6 − 1)! = 5! = 120 ways.
Mathematics for Grade XII 84
1. How many different letter arrangements can be constructed from the
following words:
a. REWARD
b. COMPUTER
c. TECHNOLOGY
d. MATHEMATICS
2. How many four-digit numbers can be arranged from the following
numbers:
a. 1, 2, 3, and 4 b. 1, 2, 3, 4, 5, 6, 7, 8, and 9
3. How many six-digit numbers can be arranged from the following numbers:
a. 2, 2, 3, 3, 3, and 4 b. 3, 5, 5, 7, 7, and 9
4. How many four-digit numbers can be constructed from the numbers: 2, 4,
5, and 7 if the numbers are not visible by 5.
5. Nine different trees are to be planted in a circular fashion. In how many
ways can the trees be planted?
6. There are 3 boys and 5 girls. In how many ways can they sit next to each
other if:
a. they can sit in any position,
b. boys and girls must be separated, such that a boys sits next to a girl.
7. A library owns 3 different books in English and 2 books in German.
a. If there are 5 places, determine the number of possibilities the books
can be arranged without having books of the same language sitting sideby-side.
b. If there are only 4 places, determine the number of possibilities of
arranging the books.
8. Find the number of possibilities in seating arrangements of ݊ couples
around a circular table, if:
a. the men and women sit in an alternating order,
b. every woman sits next to her own husband.
Classroom Activities 3
Mathematics for Grade XII 85
D. Combination
Definition
The combination of a group of objects is the number of arrangements
of the objects without taking into account the order in the
arrangements.
1. Combination of objects from different objects
Situation : There are ݊ different objects
Problem : Determining the number unorderd arrangements from
available objects
ܥ or ܥ : Notation
ܥ
= 1
Example 10
At the ABBS High School there are 10 high-achieving students. A math
teacher will select 10 out of those 10 students to participate in a national
math competition. How many student compositions can be considered by
the math teacher?
Solution
This problem is a problem of determining the combination of 10 objects
out of 10 objects because the order of selected students is out of concern,
hence:
ଵܥ
ଵ = 1
Thus, the math teacher can only consider 1 combination of students.
2. Combination of objects from different objects, ≥
Situation : There are ݊ objects which are different one from another
Problem : Determining the number unorderd arrangements which
consist of ݇ out of the available ݊ objects, ݇ ≤ ݊
ܥ ,ܥ : Notation
, or ቀ
݊
݇
ቁ
ܥ
=
1
݇!
∙ ܲ
=
1
݇!
∙
݊!
(݊ − ݇)!
=
݊!
݇! (݊ − ݇)!
Mathematics for Grade XII 86
Example 11
From the top grade XII students at the “ABIDIN” High School, three will
be selected to represent the school in an academics competition in the
Jakarta Province. If the 8 students have equal performances, in how many
ways can the selection be made?
Solution
This problem ia a problem of selecting 3 objects out of 8 objects, regardless
of their order. So the number of possible selection is
ܥ
=
଼!
ଷ!(଼ିଷ)!
=
଼!
ଷ!ହ!
=
଼∙∙
ଷ∙ଶ∙ଵ
= 56 on or combination problems.
a. The way 11 person line up to buy concert tickets
b. Choosing 2 pizza topping out of the available 12 toppings
c. Choosing 5 out of 10 basketball players
1. Determine the value of ݊ in the following equations:
a. ݊ + ܥସ
ହ = ܥଶ
b. ݊ + ܥଶ
ଷ = 2 × ܥଵ
2. A basketball coach will chose 5 out of 10 prepared players. If the 10
players can play in any position, how many ways are there to select the
players?
3. Consider 20 points where no three or more points lie on a straight line.
How many straight lines can be drawn through 2 points?
4. A farmer buys 2 goats, 3 chicken, and 4 geese from his neighbour who
owns 4 goats, 7 chicken, and 5 geese. In how many ways can the farmer
make his choices?
5. In how many ways can 9 persons be divided into 3 groups consisting of
4, 3, and 2 persons?
6. From 8 balls of different colors, 2 balls and 3 balls are selected in order.
How many possible selsctions are there?
Classroom Activities 4
Mathematics for Grade XII 87
7. In the Indonesian alphabet there are 26 letters including 5 vowels.
Determine:
a. The number of letter arrangements that can be made which consist
of 3 different consonants and 2 different vowels.
b. The number of letter arragements like in part a, but which consist the
letter b.
c. The number of letter arrangements like in part a, but which consist
the letter b and e
E. Binomial Newton
The algebraic forms for the factorization of two-term summations which are
raised to a certain power are
(ܽ + ܾ)
ଶ = ܽ
ଶ + 2ܾܽ + ܾ
ଶ
(ܽ + ܾ)
ଷ = ܽ
ଷ + 3ܽ
ଶܾ + 3ܾܽଶ + ܾ
ଷ
are examples of binomial distribution (binomial expansion) because we
expand 2 terms, namely ܽ and ܾ. the binomial expansion above can also be
written as:
(ܽ + ܾ)
= ቀ݊
݅
ቁ ܽ
ିଵܾ
ୀ
This form is the general form of the binomial Newton.
Example 12
Expand the form
The resulting coefficients can be (ݔ − 3ݕ(
ହ
!
Solution
(ݕ3 − ݔ)
ହ = ቀ
5
0
ݔ ቁ
ହ(−3ݕ(
+ ቀ
5
1
ݔ ቁ
ସ(−3ݕ(
ଵ + ቀ
5
2
ݔ ቁ
ଷ(−3ݕ(
ଶ+
ቀ
5
3
ݔ ቁ
ଶ(−3ݕ(
ଷ + ቀ
5
4
ݔ ቁ
ସ(−3ݕ(
ଵ + ቀ
5
5
ݔ ቁ
(ݕ3(−
ହ
ݔ ∙ 1=
ହ
∙ 1 + 5 ∙ ݔ
ସ
ݔ ∙ 10) + ݕ3∙ (−
ଷ
ݕ9∙
ଶ + 10 ∙ ݔ
ଶ
ݕ27∙ (−
ଷ)
ݕ81 ∙ ݔ ∙ 5+
ସ + 1 ∙ 1 ∙ (−243ݕ
ହ)
ݔ =
ହ − 15ݔ
ݔ90 + ݕସ
ݕଷ
ଶ − 270ݔ
ݕଶ
ଷ + 405ݕݔସ−243ݕ
ହ
Mathematics for Grade XII 88
Determining coefficient of binomial Newton (ܽݔܾ + ݕ(
can determined
by the following formula:
Coefficient of ݔ
ݕ ∙ ି
ܥ =
ܽ
ି ∙ ܾ
Example 13
Determine the coefficient of ݔ
ଶ
ݕ ∙
ଷ
in form (3ݔ − 2ݕ(
ହ
.
Solution
Coefficient of ݔ
ଶ
ݕ ∙
ଷ = ܥଷ
ହ3
ଶ
∙ (−2)ଷ
= 10 ∙ 9 ∙ (−8)
= −720
1. Expand the following binomials.
(ݕ − ݔ2. (a
ହ
b. ቀݔ+
ଶ
௫
ቁ
2. Determine the coefficient of:
ݔ .a
ݕଷ
ଶ
in form ቀ2ݔ−
ଵ
ଶ
ቁݕ
ହ b. ݔ
ଷ
in form (1 + ݔ + 2ݔ
ଶ)
ଷ
3. Determine ݇ and ݊, if:
(ݔ݇ + 1(
= 1 + 4√15ݔ + 60ݔ
ଶ
Classroom Activities 5
Mathematics for Grade XII 89
CHAPTER REVIEW EXERCISE
1. Given the numbers 0, 1, 2, 3, 4, 5. The total numbers that can be constructed from
the given numbers between 1000 and 5000 and has no repeatable numbers is…
A. 240 D. 360
B. 256 E. 480
C. 300
2. Three digit numbers larger than 400 are constructed from the numbers 2, 3, 4, 5,
6, and 7. The number of possibilities is … .
A. 100 D. 144
B. 108 E. 216
C. 120
3. Given 7
( 1)!
( 2)!
n
n
, the value of n is … .
A. 1 D. 7
B. 2 E. 9
C. 5
4. The number of possibilities in selecting the president, treasure, and secretary out
of 10 candidates is … .
A. 720 D. 10
B. 70 E. 9
C. 30
5. Five different books will be arranged. The number of way to arrange these books
is … .
A. 6 D. 120
B. 20 E. 240
C. 60
6. The number of different letter arrangements that can be made from the letters in
“CHOICE” is … .
A. 6 D. 360
B. 60 E. 720
C. 120
7. The total ways of sitting arrangements of six persons in a meeting at a circular
table is … .
A. 6 D. 60
B. 12 E. 120
C. 30
8. If 1
4
2
5 2
n n C C and n 5, the value of n is … .
A. 8 D. 11
B. 9 E. 12
Mathematics for Grade XII 90
C. 10
9. The football coach at ABBS has 10 players who can play in any position. The
number of ways to arrange different 6-person team that the coach can consider
is … .
A. 10 D. 1260
B. 60 E. 5040
C. 210
10. The group consists of 4 persons will be chosen from 4 boys and 7 girls. If in this
group should be least 2 girls, the number of ways to choose is … .
A. 672 D. 112
B. 330 E. 27
C. 301
11. The coefficient of 4 3
x y in the binomial 7
(3x y) is … .
A. 567 D. 2635
B. 1560 E. 2835
C. 1701
12. The coefficient of 4
x in the binomial 5
(32x) is … .
A. 16 D. 144
B. 48 E. 240
C. 96
Mathematics for Grade XII 91
Probability ii
Figure 4.1 Chance to be the winner (Furqaan Hanafi)
Wheter we reaize it or not,
probabilities and uncertainties play important
roles in our lives. In
preparingfor a test, we
make measurements the
chances a certain topic will
appear on the test. When
we going out, sometimes
we make an estimation
whether or not it will rain to
decide whether we should
bring a raincoat.
Such decisions frequently must be made among uncertainties. Analysis of
estimations concerning the probabilities that an event will take place is studiesd in
probability theories. In mathematics, the word probability is used to estimate how
big is the chance that a certain event will occur.
Bagi orang laki-laki ada hak bagian dari harta peninggalan ibu-bapa dan kerabatnya,
dan bagi orang wanita ada hak bagian (pula) dari harta peninggalan ibu-bapa dan
kerabatnya, baik sedikit atau banyak menurut bahagian yang telah ditetapkan. (Q.S.
An Nisa’ ayat 7)
Chapter 4
What is the purpose of lesson?
3.4 Mendeskripsikan dan menentukan peluang kejadian majemuk
(peluang kejadian-kejadian saling bebas, saling lepas, dan kejadian
bersyarat) dari suatu percobaan acak
4.4 Menyelesaikan masalah yang berkaitan dengan peluang kejadian
majemuk (peluang, kejadian-kejadian saling bebas, saling lepas, dan
kejadian bersyarat)
Mathematics for Grade XII 92
A. Probability of an Event
The probability of the occurence of event ܧ is denoted by ܲ(ܧ .(
Theorem
If the sample space ࡿ consists of identical sample points, such that each has
the same chance and that ࡱ is an expected event, then ࡼ)ࡱ= (
(ࡱ)
(ࡿ)
,
Where )ࡱ = (the number of members in ࡱ and )ࡿ = (the number of
elements in the sample space.
Therefore, the range of probabilities of the event is between 0 and 1. The
larger the probability (closer to 1), the more likely an event is to occur.
Example 1
A dice is tossed once. Determine:
a. the sample space of the experiment and the number of the sample space
members,
b. the probability of getting odd scores,
c. the probability of getting scores less than 5.
Solution
a. ܵ = {1,2,3,4,5,6}, ݊(ܵ) = 6
b. ܧ = dice landing on odd scores
= {1,3,5}
3) = ܧ)݊
= (ܧ)ܲ
(ா)
(ௌ)
=
ଷ
=
ଵ
ଶ
c. ܧଵ = dice landing on scores less than 5
= {1,2,3,4}
݊(ܧଵ) = 4
= (ଵܧ)ܲ
(ாభ(
(ࡿ)
=
ସ
=
ଶ
ଷ
Mathematics for Grade XII 93
Example 2
In a box there are 4 red marbles, 3 blue marbles, and 2 white marbles. Three
marbles are taken together randomly. Determine the probability that the
marbles taken are:
a. 2 red and 1 blue
b. 1 red, 1 blue, and 1 white
c. all red
Solution
In total there are 9 marbles. Three marbles can be taken from the box is
ଷܥ
ଽ =
ଽ!
!ଷ!
= 84 ways. So the members of the universe in taking of 3 out of 9
marbles are ݊(ܵ) = 84.
a. In the box there are 4 red marbles. Two red marbles can be taken from the
box in ܥଶ
ସ = 6 ways.
In the box there are 3 blue marbles. One blue marble can be taken in ܥଵ
ଷ =
3 ways.
So, two red marbles and one blue marble can be taken from the box in ܥଶ
ସ
∙
ଵܥ
ଷ = 6 ∙ 3 = 18 ways.
Hence ܲ(2ݎ ,1ܾ) =
ଵ଼
଼ସ
=
ଷ
ଵସ
.
b. The probability of taking 1 red, 1 blue, and 1 white is:
ܲ(1ݎ ,1ܾ, 1ݓ= (
భ
ర
∙భ
య
∙భ
మ
య
వ =
ସ∙ଷ∙ଶ
଼ସ
=
ଶ
c. The probability of taking 3 red marbles is:
= (ݎ3ܲ(
య
ర
య
వ =
ସ
଼ସ
=
ଵ
ଶଵ
Mathematics for Grade XII 94
1. In a trial tossing 2 red and 1 white dice once, what is the probability of
getting:
a. a total score of 9 out of the two dice,
b. an odd score on the red dice
c. a score on the white dice which is a factor of the scores on the red dice
2. The letters in the word “syahadatain” are written on pieces and are put
inside a box. The paper pieces are then mixed and a piece of paper is
randomly taken. What is the probability of getting a piece of paper with
the letter A in that event?
3. If 3 coins are tossed together once, what is the probability of getting:
a. all heads
b. 2 heads and 1 tail
c. ony 1 head
d. at least 1 head
B. Probability of the Combination of Two Mutually
Exclusive Events
Suppose that ܧଵ and ܧଶ are two events in the same experiment.
= (ଶܧ ∪ ଵܧ)ܲ
(ாభ∪ாమ
)
(ௌ)
= (ଶܧ ∪ ଵܧ)ܲ
(ாభ∪ாమ
)
(ௌ)
=
(ாభ)ା(ாమ
)ି(ாభ∩ாమ
)
(ௌ)
=
(ாభ
)
(ௌ)
+
(ாమ
)
(ௌ)
−
(ாభ∩ாమ
)
(ௌ)
ܲ(ܧଵ ∪ ܧଶ) = ܲ(ܧଵ) + ܲ(ܧଶ) − ܲ(ܧଵ ∩ ܧଶ) ...... (1)
Two event ܧଵ and ܧଶ often times do not intersect, i.e. ܧଵ ∩ ܧଶ = ∅. If the
intersection between the two events is a null set, it is said the two events are
mutually exlclusive (disjoint). It can also be said that events ܧଵ and ܧଶ do
not occur simulatneously.
Suppose in the toss of dice, events ܧଵ and ܧଶ are defined as follows.
ܧଵ = the occurence of prime numbers {2, 3, 5}
Classroom Activities 1
Mathematics for Grade XII 95
ܧଶ = the occurence of quadratic numbers {1,4}
ܧଵ ∩ ܧଶ = { }, hence ܧଵ and ܧଶ are called mutually exclusive events.
The probability of the intersection of two mutually exclusive events is
= (ଶܧ ∩ ଵܧ)ܲ
(ாభ∩ாమ
)
(ௌ)
= 0
Hence (1) becomes:
(ଶܧ)ܲ + (ଵܧ)ܲ = (ଶܧ ∩ ଵܧ)ܲ
The probability of two mutually exclusive events (disjoint) ܧଵ and ܧଶ is
.(ଶܧ)ܲ + (ଵܧ)ܲ = (ଶܧ ∩ ଵܧ)ܲ
Example 3
In a dice toss, A is the event of the occurence of prime numbers and B is the
event of the occurence of numbers that are multiples of 3. Determine the
probability of the occurence of prime numbers or numbers that are multiples
of 3.
Solution
In this experiment, ݊(ܵ) = 6.
The event of the occurence of prime numbers is ܣ} = 2,3,5} so ݊(ܣ = (3
The event of the occurence of numbers that are multiplies of 3 is ܤ} = 2,3},
so ݊(ܤ = (2.
The event of the occurence of prime numbers or numbers that are multiples
of 3 is ܣ ∪ ܤ} = 2, 3, 5, 6} so ݊(ܣ ∪ ܤ = (4
= (ܤ ∪ ܣ)ܲ ,Hence
(∪)
(ௌ)
=
ସ
=
ଶ
ଷ
.
Example 4
A bag contains 10 red marbles, 18 green marbles, and 22 yellow marbles.
If a marble is taken randomly from the bag, determine the probability tht
the marble is red or yellow.
Solution
In this experiment, ݊(ܵ) = 50
Let: A = the occurence of a red marble
B = the occurence of a green marble
C = the occurence of yellow marble
Mathematics for Grade XII 96
Events A, B, and C are mutually exclusive.
= (ܣ)ܲ
()
(ௌ)
=
ଵ
ହ
= (ܥ)ܲ and
()
(ௌ)
=
ଶଶ
ହ
(ܥ)ܲ + (ܣ)ܲ = (ܥ ∪ ܣ)ܲ ,Hence
=
ଵ
ହ
+
ଶଶ
ହ
=
ଷଶ
ହ
=
ଵ
ଶହ
1. In a trial of tossing dice at the same time, determine:
a. The probability of getting a score of 3 on one dice or a total score of
5from both dice
b. The probability of getting a total score of more than 3
2. One card is taken from a deck of cards. Determine the probability of
getting an ace or a king
3. Two ball are to be taken randomly from a box containing of 3 purple balls,
6 green balls and 1 yellow ball. Determine the probability of getting.
a. two green balls
b. two yellow balls
c. a yellow ball or a purple ball
C. The Probability of Two Independent Events
The probability events mean that one event does not influence another one,
or one event does not depend on another one. For example, in the toss of a
coin and dice at the same time.
ܧଵ = the occurence of a head = {ܪ{
ܧଶ = the occurence of even numbers = {2,4,6}
In the toss described above, the occurence of a head does not influence the
occurence of even numbers, and vice versa. All probabilities in the toss of a
coin and a dice at the same time are given in the table below.
Classroom Activities 2
Mathematics for Grade XII 97
Dice
Coin
1 2 3 4 5 6
Head (H) (ܪ ,1) (ܪ ,2) (ܪ ,3) (ܪ ,4) (ܪ ,5) (ܪ ,6)
Tail (T) (ܶ, 1) (ܶ, 2) (ܶ, 3) (ܶ, 4) (ܶ, 5) (ܶ, 6)
From the table we see that ݊(ܵ) = 12 and ݊(ܧଵ ∩ ܧଶ) = 3.
Hence, ܲ(ܧଵ ∩ ܧଶ) = (ாభ∩ாమ)
(ௌ)
=
ଷ
ଵଶ
=
ଵ
ସ
Besides the method described above, we can also obtain ܲ(ܧଵ ∩ ܧଶ) by
multiplying ܲ(ܧଵ) with ܲ(ܧଶ).
݊(ܧଵ) = 1 and ݊(ܵாభ
) = 2, so ܲ(ܧଵ) = (ாభ)
(ௌಶభ
)
=
ଵ
ଶ
݊(ܧଶ) = 3 and ݊(ܵாమ
) = 6, so ܲ(ܧଶ) = (ாమ)
(ௌಶమ
)
=
ଷ
(ଶܧ)ܲ ∙ (ଵܧ)ܲ = (ଶܧ ∩ ଵܧ)ܲ
=
ଵ
ଶ
∙
ଷ
=
ଵ
ସ
Based on the description, we can conclude that:
Two events ܧଵ and ܧଶ are called independent if and only if ܲ(ܧଵ ∩ ܧଶ) = ܲ(ܧଵ) ∙
(ଶܧ)ܲ
Example 5
Two coins are tossed at the same time. Let ܣ be the evet of the occurence of head
on the first coin and ܤ the event of the occurence of head on the second coin.
Determine the propability of event ܣ and event ܤ .
Solution
Because there are two different coins, the event on the first coin does not influence
what happens on the other, so ܣ and ܤ are independent.
(ܤ)ܲ ∙ (ܣ)ܲ = (ܤܣ)ܲ
=
ଵ
ଶ
∙
ଵ
ଶ
=
ଵ
ସ
Hence, the probability of event ܣ and event ܤ is ଵ
ସ
.
Mathematics for Grade XII 98
Example 6
Two dice are tossed together, one is red and the other is green. If ܣ is the event of
the occurence of 2 on the red dice and ܤ is the event that the total score on both is
5, are event ܣ and event ܤ independent?
Solution
= (ܣ)ܲ
ଵ
= (ܤ)ܲ and
ସ
ଷ
= (ܤ ∩ ܣ)ܲ,{(3,2 = {(ܤ ∩ ܣ
ଵ
ଷ
= (ܤ)ܲ ∙ (ܣ)ܲ
ଵ
∙
ସ
ଷ
=
ସ
ଶଵ
(ܤ ∩ ܣ)ܲ =/
Hence, event ܣ and event ܤ are not independent.
Example 7
The probability that ܣ will live for another 20 years is 0.75 whereas the probability
of ܤ is 0.82. determine the probability that both ܣ and ܤ will live for another 20
years.
Solution
The events of ܣ and ܤ living for another 20 years are two independent events.
Thus, the probability that both ܣ and ܤ will live for another 20 years is (0.75) x
(0.82) = 0.615.
Mathematics for Grade XII 99
1. A box contains 6 red balls and 4 green balls. Two balls are taken from the box,
one at a time with no replaclement. Determine the probability that the first ball
is red and the second is green.
2. A bag contains 5 red marbles, 3 white marbles, and 2 green marbles, Two
marbles are taken at once from the bag. Determine the probability of gettingone
red marble and one green marble.
3. In the toss of a white dice and a red dice together, determine the probability of
getting 2 on the red dice and on the white dice.
4. Box I contains 3 red balls and 2 white balls, whreas box II contains 5 red balls
and 3 black balls. If one ball is taken from each box, determine the probability
of getting:
a. a red ball from box I and a black ball from box II,
b. a white ball from box I and a black ball from box II,
c. a red ball from each box.
D. The Probability of Conditional Events
The probability of an event in a trial where is additional information given
regarding the results is called the probability of conditional events. A proble
on conditional probability is usually stated in this way: “What is the probability
of an event . . . given . . .”
The probability of event A given that B has occured is denoted by the symbol
.(ܤ|ܣ)ܲ
Consider the following Venn diagram.
What is the probability of A given that B has occured? Because B has occured
our sample space is the circle B, and part of A that is also in B is only ܣ ∩ ܤ .
Hence
= (ܤ|ܣ)ܲ
(∩)
()
=
(ಲಳ)
(ೄ)
(ಳ)
(ೄ)
=
(∩)
()
Classroom Activities 3
Mathematics for Grade XII 100
= (ܤ|ܣ)ܲ
(ܤ ∩ ܣ)ܲ
(ܤ)ܲ
The probability of A given that B has occured is
= (|)ࡼ
( ∩ )ࡼ
()ࡼ
Example 8
A card is taken from eight identical cards that have been numbered 1, 2, ..., 8.
What is the probability of getting a prime-numbered card given that the selected
cars has an odd number?
Solution
By applying the conditional probabilitythat we have just indroduced.
Let
ܧଵ = the occurence of getting a prime-numbered card = {2, 3, 5, 7}
ܧଶ = the occurence of getting an odd-numbered card = {1, 3, 5, 7}
Then ܧଵ ∩ ܧଶ = {3, 5, 7}
= (ଶܧ ∩ ଵܧ)ܲ
(ாభ∩ாమ
)
(ாభ)
=
ଷ
ସ
Hence, the probability of getting a prime-numbered card given that the selected
card has an odd numer is ଷ
ସ
.
From the equation ܲ(ܣ|ܤ= (
(∩)
()
, we can derive the equation
(ܤ|ܣ)ܲ ∙ (ܤ)ܲ = (ܤ ∩ ܣ)ܲ
Example 9
A box contains 5 red balls and 3 green balls. A ball is randomly taken from the
box and the another ball is taken while the first ball is not returned into the box.
Dteermine the probability that the two balls taken are green.
Solution
Let
ܧଵ = the occurence of a green ball in the first selection
ܧଶ = the occurence of a green ball in the second selection
The probability of ܧଵ and ܧଶ = ܲ(ܧଵ ∩ ܧଶ) = ܲ(ܧଵ) ∙ ܲ(ܧଵ|ܧଶ)
Mathematics for Grade XII 101
Because ܲ(ܧଵ) =
ଷ
଼
and ܲ(ܧଵ|ܧଶ) =
ଶ
, then
= (ଶܧ ∩ ଵܧ)ܲ
ଷ
଼
∙
ଶ
=
ହ
=
ଷ
ଶ଼
Hence, the probability that the two balls taken are green is ଷ
ଶ଼
.
This problem can also be solved with the multiplication rules as follows.
First Selection Second Selection Result Probability
ହ
଼
red
ଷ
଼
green
ଷ
green
ସ
red
(red, green) 5
8
∙
3
7
(red, red) 5
8
∙
4
7
ଶ
green
ହ
red
(green, green) 3
8
∙
2
7
(green, red) 3
8
∙
5
7
Example 10
A bag contains 3 red marbles, 2 green marbles, and 1 yellow marble. Two
marbles are taken at once. Determine the probability that one marble is red and
the other is yellow.
Solution
Method 1
The selection of two marbles can happen i two ways: the first is red and the
second yellow, and the first is yellow and the second red.
On the first possibility: ܲଵ(݉) =
ଷ
and ܲଶ(݇|݉) =
ଵ
ହ
On the second possibility: ܲଵ(݇) =
ଵ
and ܲଶ(݉|݇) =
ଷ
ହ
ܲ(݉ ∩ ݇) = ሾܲଵ(݉) ∙ ܲଶ(݇|݉)ሿ + ሾܲଵ(݇) ∙ ܲଶ(݉|݇)ሿ
= ቀ
ଷ
∙
ଵ
ହ
ቁ + ቀ
ଵ
∙
ଷ
ହ
ቁ
=
ଷ
ଷ
+
ଷ
ଷ
=
ଵ
ହ
Hence, the probability of getting a red marble and a yellow marble is ଵ
ହ
.
Mathematics for Grade XII 102
Method 2
We can also solve this problem by using the rules of combination, as follows:
The number of ways to select a red marble and a green marble is ቀ
3
1
ቁ ቀ1
2
ቁ
The number of ways to select 2 marbles out of 6 marbles is ቀ
6
2
ቁ.
Hence, the probability of selecting a red marble and a green marble at once is
ቀ
ଷ
ଵ
ቁቀଵ
ଵ
ቁ
ቀ
ଶ
ቁ
=
ଷ∙ଵ
ଵହ
=
ଵ
ହ
.
Example 11
There are 5 people who will sit in a row. Two of them are Rini and Rina, they
are twins. If they don't want to sit next to each other, then the probability is ...
.
A. ଵ
ହ
C. ଷ
ହ
E. 1
B. ଶ
ହ
D. ସ
ହ
Solution
There are 5 people who will sit in a row. To find probability so that 2 of them
do not sit next to each other can be easier if you use the complement.
The complement means that Rina and Rini sit next to each other, so that Rina
and Rini are considered one unit. So the number of ways they can sit is 4!.
While Rina and Rini are free to move on the right or left, so that there are 2
sitting positions for Rina and Rini.
So the number of ways 5 people sit in a row with Rina and Rini next to each
other is:
ܣ)݊
) = 4! 2! = 4.3.2.1.2.1 = 48
While the universe is:
݊(ܵ) = 5! = 5.4.3.2.1 = 120
So the probability that Rina and Rini are next to each other is
ܣ)ܲ
) =
ܣ)݊
)
݊(ܵ)
=
48
120 =
2
5
So the probability that Rini and Rina are not next to each other while sitting
ܣ)ܲ − 1) = ܣ)ܲ is
) = 1 − ଶ
ହ
=
ଷ
ହ
(C).
REMEMBER
ቀ
= ቁ
=
!
! !( − )
Mathematics for Grade XII 103
Example 12
A dice is unequal, the probability of getting number 4 is one-fourth of the other
totals. The probability of number 2 and 3 appearing are each one-third of the
total other totals. If it is tossed 3 times, then the probability of getting a number
less than four twice is ... .
A. 0,576 C. 0,324 E. 0,108
B. 0,467 D. 0,216
Solution
1) The probability that the number 1 appears is a quarter of the other totals
P(1) =
1
4
൫1 − P(1)൯ =
1
4
−
1
4
P(1)
⟺
5
4
P(1) =
1
4
⟺ P(1) =
1
5
2) The probability that the numbers 2 and 3 will appear is one-third of the
other totals
P(2) =
1
3
൫1 − P(2)൯ =
1
3
−
1
3
P(2)
⟺
4
3
P(2) =
1
3
⟺ P(2) =
1
4
P(3) =
1
3
൫1 − P(3)൯ =
1
3
−
1
3
P(3)
⟺
4
3
P(3) =
1
3
⟺ P(3) =
1
4
3) Then the probability of getting a number less than four
P(1 or 2 or 3) = P(1) + P(2) + P(3) =
1
5
+
1
4
+
1
4
=
7
10
Mathematics for Grade XII 104
4) If you throw 3 times, then the probability of getting a number less than 4
is twice =
ଶ
ଷ
. P(1 or 2 or 3) =
ଶ
ଷ
.
ଵ
= 0,467 (B).
Example 13
A die contains 6 irregular faces, each side is numbered 1,2,3,4,5 and 6. If the
die is rolled, it will fall on a certain side. If P(n) is the probability value of the
object falling on the number n side and the probability value is ܲ(݊) =
௫
ଶ
షభ,
then the value of x is ... .
A. ଵ଼
ଷଷ
C. ଶ
ହ
E. ଷଷ
ଵ
B. ଶ
ଵ
D. ଷଶ
ଷ
Solution
Given ܲ(݊) =
௫
ଶ
షభ
P(n) is the probability of getting side numbered n of the die.
Since there are 6 numbers, then:
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
⟺
ݔ
2
ଵିଵ +
ݔ
2
ଶିଵ +
ݔ
2
ଷିଵ +
ݔ
2
ସିଵ +
ݔ
2
ହିଵ +
ݔ
2
ିଵ = 1
⟺
ݔ
1
+
ݔ
2
+
ݔ
4
+
ݔ
8
+
ݔ
16 +
ݔ
32 = 1
⟺
ݔ + ݔ2 + ݔ4 + ݔ8 + ݔ16 + ݔ32
32 = 1
= ݔ ⟺ 32 = ݔ63⟺
ଷଶ
ଷ
(D)
Mathematics for Grade XII 105
1. A box contains 5 red balls, 3 white balls, and 2 green balls. A ball is
randomly taken from the box. Determine the probability that the ball taken
is not green.
2. Two dice are tossed together once. Determine the probability that the sum
of the two dice are:
a. greather than 5 b. less than 11
3. A bag contains 4 white balls and 5 yellow balls. If 2 balls are drawn from
the bag, find the probability that the drawn balls are yellow if:
a. the first balls is replaced before the second ball is drawn,
b. if the first ball is not replaced
4. Three coins are tossed together. Determine the probability of getting:
a. two tails b. at least 1 head
5. The probability that team A will win a basketball game from team B is 0,4.
In a series of 5 games, what number of games is A most likely to win? Is
A more likely to win exactly 1 or exactly 3 games?
6. The probability that the Red Java will win their conference championship
is ଷ
଼
, and the probability that the Silver Sumatra will win the same
conference championship is ଵ
ସ
. Find the probability that one or the other of
these teams will be the champion.
Classroom Activities 4
Mathematics for Grade XII 106
CHAPTER REVIEW EXERCISE
1. Box A contains 6 red balls and 4 yellow balls. Box B contains 5 red balls and 7
yellow balls. If a ball is taken from each box, the probability of getting a red ball
from box A and a yellow ball from box B is … .
A.
6
1 D.
30
7
B.
4
1
E.
20
7
C.
10
3
2. If two dices are tossed together once, the probability of getting a total point of
five or even is … .
A.
9
1
D.
3
2
B.
2
1
E.
12
10
C.
18
11
3. A student has to solve 8 from 10 questions, but number 1 up to 4 have to be
solved. The number of choice to solve that questions is … .
A. 10 D. 25
B. 15 E. 30
C. 20
4. A box which contains 8 red marbles and 10 white marbles will be taken once
randomly. The probability of getting 2 white marbles is … .
A.
153
20 D.
153
56
B.
153
28 E.
153
90
C.
153
45
5. The box contains 4 yellow balls and 6 blue balls. If 2 balls are taken once
randomly the the probability of getting the same colour balls is … .
A.
15
2
D.
15
7
B.
15
3
E.
15
8
C.
15
5
Mathematics for Grade XII 107
6. Two dices are tossed together once. The probability of getting a total point of 9
or 11 is … .
A.
2
1
D.
8
1
B.
4
1
E.
12
1
C.
6
1
7. A box contains 10 balls which have been numbered from 1 to 10. Two balls are
taken repeatedly 80 times from the box. The expectation frequency of getting an
even total value from the balls is … .
A. 20 times D. 50 times
B. 30 times E. 60 times
C. 40 times
8. A box contains 6 red balls and 4 white balls. Two balls are taken randomly
without replacement. If the trial is conducted 90 times, the expectated frequency
of getting a red ball and a white ball is … .
A. 72 times D. 24 times
B. 48 times E. 12 times
C. 45 times
Mathematics for Grade XII viii
BIBLIOGRAPHY
Bornok Sinaga, dkk. 2014. Matematika SMA/MA SMK/MAK Kelas XI. Jakarta: Pusat
Kurikulum dan Perbukuan, Balitbang, Kemdikbud
Dobbs, Steve and Miller, Jane. 2012. Statistics 1. Cambridge:Cambridge University Press
Miyanto, dkk. 2015. Matematika (Peminatan Matematika dan Ilmu-Ilmu Alam) SMA/MA
Kelas XII. Klaten: Intan Pariwara
Probability and Statistics. ConnectED.mcgraw-hill.com
Sri Kurnianingsih, Kuntarti, Sulistiyono. 2010. Mathematics for Senior High School Grade
XI. Jakarta:Erlangga
Sri Kurnianingsih, Kuntarti, Sulistiyono. 2012. Mathematics for Senior High School Grade
XII. Jakarta:Erlangga
Sukino. 2014. Matematika untuk SMA/MA Kelas XI. Jakarta:Erlangga
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3A
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https://www.google.co.id/search?q=bismillah&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjLos
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aWIBg&q=kaligrafi+alhamdulillahi+rabbil+alamin&oq=alhamdulillahi+kaligrafi&gs_l=img.
Mathematics for Grade XII ix
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