X- Maths CH 3 (E) Full Notes

Chapter 3
Mathematics Of Chance

Prepared by
Cecilia Joseph,
HST,
St. John De Britto’s,A.I.H.S,
Fortkochi

Mathematics Of Chance

Online Class – X − 28 27 / 08 / 2021

3. Mathematics Of Chance – Class 1

To view class

Possibilities as numbers

Consider following situations
* When a coin is tossed we cannot predict early whether the

result of the toss will be “head” or “tail” . We can only
assume the result.
* While throwing a die we cannot predict early which
number is coming upward. We can only assume the result.

Mathematical analysis of cases
where the result cannot be predicted

early is discussing in this unit

Activity 1

There are 9 orange balls and one rose ball in a box .

If you pick a ball from the box without looking, it is most

likely to be of which colour ?

Ans)

In the box the number of orange balls is

more than the number of rose ball. 9 Orange balls
∴

The probability of getting an 1 Rose ball
orange ball is larger .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance

Activity 2

There are 8 orange balls and 2 rose balls in a box .

If you pick a ball from the box without looking, it is most

likely to be of which colour ?

Ans)

In the box the number of orange balls is more

than the number of rose balls. 8 Orange balls
∴ The probability of getting an orange ball 2 Rose balls

is larger .

Activity 3

There are 5 orange balls and 5 rose balls in a box .

If you pick a ball from the box without looking, it is most

likely to be of which colour ?

Ans)

In the box the number of orange balls and

number of rose balls are same . 5 Orange balls
∴ The probability of getting an orange ball
5 Rose balls
or a rose ball are same .

Activity 4
a)There are 5 orange balls and 5 rose balls in a box and

6 orange balls and 4 rose balls in another box .
One has to choose a box and pick a ball (without looking ) .
If it is orange, he wins. Which box is the better choice?
Ans)

5 Orange balls 6 Orange balls

5 Rose balls 4 Rose balls

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance

Number of balls in first box = 5 + 5 = 10
Number of balls in second box = 6 + 4 = 10
Both box contain same number of balls.
Second box contains more orange balls, so we have a
greater probability of getting an orange ball from the
second box.
∴ Second box is the better choice.

b)There are 6 orange balls and 5 rose balls in a box and
5 orange balls and 4 rose balls in another box .
One has to choose a box and pick a ball (without looking ) .
If it is orange, he wins. Which box is the better choice?

Ans)

6 Orange balls 5 Orange balls

5 Rose balls 4 Rose balls

Number of balls in first box = 6 + 5 = 11
Number of balls in second box = 5 + 4 = 9

Probability of getting an orange ball from the first box = 6
11
5
Probability of getting an orange ball from the second box = 9

Comparing 6 & 5 we can see 6 = 6x9 = 54
11 9 11 11 x 9 99

5 is the larger. 5 = 5 x 11 = 55
9 9 9 x 11 99

Probability of getting an orange ball from 5 6
9 11
the second box is larger. ∴ larger than

∴ Second box is the better choice.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance

The probability of something we have to find is
how much part of the total number of results
to the number of results favourable to it
That is
Number of favourable results

Probability = Total number of results

Activity 5

There are 5 black and 4 white balls in a box . A ball is taken

from it without looking

a) What is the probability of it being black ?

b) What is the probability of it being white ?

Ans)

Total number of results = 5 + 4 = 9 5
9
a) Probability of the ball being black = 5 Black balls

4 White balls

b) Probability of the ball being white = 4
9

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 1

Online Class – X − 29 31 / 08 / 2021

3. Mathematics Of Chance – Class 2

To view class

Activity 1
There are 7 red and 5 blue balls in a bag , 9 red and 7 blue balls
in another .
a) What is the probability of getting a red ball from the first

bag ?
b) What is the probability of getting a red ball from the second

bag ?
c) If all the balls are put in a single bag ,what is the probability

of getting a red ball from it ?
Ans)

7 red balls 9 red balls
5 blue balls 7 blue balls

a) Number of balls in the first bag = 7 + 5 = 12 7
12
Probability of getting a red ball from first bag =

b) Number of balls in the second bag = 9 + 7 = 16 9
16
Probability of getting a red ball from second bag =

c) Balls are put in a single bag

So, total number of balls in the bag = 12 + 16 = 28

Total number of red balls in the bag = 7 + 9 =16
16
Probability of getting a red ball from this bag = 28

= 4x4
7x4
4
= 7

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 2

Activity 2

Numbers 1 to 25 are written on slips of paper and put in a

box . A slip is to be drawn from it ,

a) What is the probability of getting an even number ?

b) What is the probability of getting an odd number ?

Ans)

a) Even numbers ‍= 2 , 4 , 6, 8 , 10 , 12 ,14 , 16 , 18 , 20 , 22 , 24

Number of favourable results = 12 12
25
Probability of getting an even number =

b) Odd numbers = 1 , 3 , 5 , 7 , 9 , 11 , 13 ,15 , 17 , 19 , 21 , 23 , 25

Number of favourable results = 13 13
25
Probability of getting an odd number =

Note:

In the above problem ‍,

Probability of getting + Probability of getting = 12 13
an even number 25 25
an odd number +

= 25 =1
25

Geometrical probability
Activity 1
A circle is divided into two equal parts .
Calculate the probability of a dot put ,
without looking , to be in the red part .

Ans) 1
2
The area of the red part is of the area of the circle .

∴ Probability of the dot falling within the red part = 1
2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 3

Activity 2
A circle is divided into four equal parts .
Calculate the probability of a dot put ,
without looking , to be in the red part .

Ans) 3
4
The area of the red part is of the area of the circle .

∴ Probability of the dot falling within the red part = 3
4

Activity 3
A circle is divided into eight equal parts .
Calculate the probability of a dot put ,
without looking , to be in the red part .

Ans) 5
8
The area of the red part is of the area of the circle .

∴ Probability of the dot falling within the red part = 5
8

Activity 4
A multicoloured disc spins on a board.
i) What is the probability of getting

red against the arrow when it stops ?
ii) Calculate the probabilities of other

colours?

Ans)

Spinning wheel has been divided into 8 equal parts.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 4

i) Probability of getting red = 4
8
4x1
= 4x2

= 1
2

ii) Probability of getting yellow = 3
8

Probability of getting green = 1
8

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 1

Online Class – X − 30 02 / 09 / 2021

3. Mathematics Of Chance – Class 3

To view class

Activity 1

A cardboard rectangle is cut out and the midpoint of one side

is joined to the ends of the opposite sides

to make a triangle. If you shut your eyes

and put a dot in this rectangle, what is the

probability that it would be within the red

triangle?

Ans)

The triangle and rectangle have the

same base and height.

Area of rectangle = bh h Height(h)
1
Area of red triangle = 2 bh

Base(b)

That is , area of the triangle is 1 the area of the rectangle .
2

∴ Probability of dot falling 1
within the red triangle
=2

Activity 2
A cardboard parallelogram is cut out and divided into two
triangles by drawing a diagonal . If
you shut your eyes and put a dot in
this parallelogram , what is the
probability that it would be within
the green triangle ?

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 2

Ans)
Diagonal of a parallelogram divide it into two equal triangles .
So their areas are equal .
ie , area of a triangle is half the area of the parallelogram .

∴ Probability of dot falling = 1
within the green triangle 2

Activity 3
In the figure midpoints of the sides of the larger
triangle are joined . If you shut your eyes and
put a dot in this figure , what is the probability
that it would be within the green triangle ?

Ans)

Here , if we fold each yellow triangle each will be exactly

aligned inside the green triangle..
1
ie , Area of the green triangle is 4 of the area of the large

triangle.

∴ Probability of dot falling = 1
within the green triangle 4

T.B Page 72
In each picture below, the explanation of the green part is
given. Calculate in each, the probability of a dot put, without
looking, to be within the green part.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 3

Q1) A square got by joining the mid
points of a bigger square.

Ans)

Diagonals of the green triangle is drawn.

Here , if we fold each yellow triangle each

will be exactly aligned inside the green

triangle. 1
2
ie, Area of the green square is of the

area of the large square.

∴ Probability of a dot put to = 4 = 1
be within the green square 8 2

Q2) A square with all vertices on a circle.

2 cm

Ans) 2√2 cm 2 cm
Area of the square = side × side
=2×2
= 4 sq . cm

Diagonal of the square = 2√2 cm 2 cm

Diagonal of the square = Diameter of the circle
= 2√2 cm

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 4

So, radius of the circle = 2√ 2 = √2 cm
2

Area of the circle = π r2 = π × √2 × √2
= 2 π sq . cm

∴ Probability of a dot put to be = 4 = 2
within the green square 2π π

Q4) A triangle got by joining alternate
vertices of a regular hexagon.

Ans)
Here , if we fold each yellow triangle each
will be exactly aligned inside the green
triangle.
So we can say sum of the areas of the
3 yellow triangles is equal to the area of
small green triangles inside the regular hexagon.

ie , Area of the green triangle is half of the area of the
regular hexagon .

∴ Probability of a dot put to be = 1
within the green triangle 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 5

Pairs
Activity 1
* Looking for a clean dress, Johny found a pair of blue pants

and three shirts, red, green and blue. “
In how many ways he can dress?

Ans)

Johny can wear dress in 3 ways.

( Blue , Red ) ( Blue , Green ) ( Blue , Blue )

* Searching again Johny found a pair of green pants also.
How many ways this can be worn with each of the three
shirts?

Ans)
Johny can wear dress in 3 more ways.

( Green , Red ) ( Green , Green ) ( Green , Blue )

Thus Johny can dress in six different ways .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 6

* What is the probability of Johny wearing shirt and pants of
the same colour?

Ans) Total no: of ways Johny can dress = 6
No: of ways of wearing same colour dress = 2

( Blue , Blue ) ( Green , Green )

Probability of wearing shirt and 2 =1
pants of the same colour =6 3

* What is the probability of Johny wearing shirt and pants of
the different colour?

Ans) Total no: of ways Johny can dress = 6
No: of ways of wearing different colour dress = 4

( Blue , Red ) ( Blue , Green ) ( Green , Red ) ( Green , Blue )

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 7

∴ Probability of wearing shirt and = 4
pants of different colour 6
2x2
= 2x3

= 2
3

Assignment

T.B Page 72

Q3) Consider a circle exactly fitting inside

a square. If we put a dot without looking

in this square ,what is the probability of

it being within the circle ?

Q5) A regular hexagon formed by two
overlapping equilateral triangles.
If we put a dot without looking in
this figure, what is the probability
of it being within the hexagon ?

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 1

Online Class – X − 31 03 / 09 / 2021

3. Mathematics Of Chance – Class 4

To view class

Answers of last class assignment
T.B Page 73
Q3) Circle exactly fitting inside a square.

Ans) r.r
If radius of circle is ‘r’
Side of the square = 2r 2r
Area of the square = 2r × 2r = 4r2
Area of the circle = πr 2

∴ Probability of a dot put to be = πr 2
within the green circle
4r2

= π
4

Q5) A regular hexagon formed by two

overlapping equilateral triangles .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 2

Ans)
By drawing diagonals we can divide green regular
hexagon into 6 equal triangles.

Now there are 12 triangles and they are
equal triangles.
( If we fold each yellow triangle each

will be exactly aligned inside the green
hexagon.)

∴ Probability of a dot put to be = 6 = 1
within the hexagon 12 2

Activity 1
If we put a dot in the picture without
looking what is the probability of it
being within the red part ?

Ans) We can see the 2 red semicircles together forms a circle .

Let, Radius of red circle = r
Area of red circle = πr 2

Radius of blue circle = 2r
Area of blue circle = π(2r) 2
= 4πr 2

∴ Probability of a dot put to be = πr2 = 1
within the red circle 4 πr 2 4

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 3

Activity 2
A box contains four paper slips numbered 1, 2, 3, 4 and
another contains two slips numbered 1, 2.
One slip is picked from each.
a) What are the possible pairs?
Ans)
Combining each number from the first box with the two
possibilities from the second,

Possible pairs are 11
(1, 1) , (1, 2) 2
(2, 1) , (2, 2)
(3, 1) , (3, 2) 3 2
(4, 1) , (4, 2) 4

Total no: of pairs = 4 × 2 = 8

( 4 x 2 = 8)

b) What is the probability of both numbers being odd ?

Ans)

Pairs in which both numbers being odd are ( 1, 1) and (3, 1).
2 1
∴ Probability of both numbers being odd = 8 = 4

c) What is the probability of both numbers being even ?

Ans)

Pairs in which both numbers being even are ( 2, 2) and (4, 2).
2 1
∴ Probability of both numbers being even = 8 = 4

d) What is the probability of one being odd and the other even?
Ans)

Pairs in which one number is odd and the other even are
(1, 2) , (2, 1) , (3, 2) , (4, 1)

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 4

∴ Probability of one being odd and other even = 4 = 1
8 2

e) What is the probability of both being the same number?

Ans)

Pairs in which both numbers being the same number are

( 1, 1) and (2, 2). 2 1
8 4
∴ Probability of both numbers being same = =

T.B Page 75
Q2) A box contains four slips numbered 1, 2, 3, 4 and another

box contains two slips numbered 1, 2. If one slip is taken
from each,
i) What is the probability of the sum of numbers being odd?
ii) What is the probability of the sum being even?
Ans) Combining each number from the first box with the two

possibilities from the second,
Possible pairs are
(1, 1) , (1, 2)
(2, 1) , (2, 2)
(3, 1) , (3, 2)
(4, 1) , (4, 2)

Total no: of pairs = 8

i) Pairs having sum of numbers odd are,

(1, 2) , (2, 1) , (3, 2) , (4, 1) 4 1
8 2
∴ Probability of the sum of numbers being odd = =

i) Pairs having sum of numbers even are,

(1, 1) , (2, 2) , (3, 1) , (4, 2) 4 1
8 2
∴ Probability of the sum of numbers being even = =

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 5

More pairs
Activity 1

One box containing ten paper slips numbered 1 to 10 and the
other box contains five paper slips numbered 1 to 5.
One slip is taken from each box, as usual.
a) What is the probability of both being odd?
b) What is the probability of both being even?
Ans)

One box contains ten paper slips numbered
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Other box contains five paper slips numbered
1, 2, 3, 4, 5

We can pair them as follows

Box 2

Box 1

Total no. of possible pairs = 10 × 5 = 50 pairs

a) Number of odd numbers in the first box = 5 (1, 3, 5, 7, 9 )
Number of odd numbers in the second box = 3 (1, 3, 5 )
No: of possible pairs in which both are odd = 5 × 3
= 15 pairs

∴ Probability of getting odd = 15 = 3
numbers from both boxes 50 10

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 6

b) Number of even numbers in the first box = 5 (2, 4, 6, 8, 10)

Number of even numbers in the second box = 2 (2, 4)

No: of possible pairs in which both are even= 5 × 2 =10 pairs

∴ Probability of getting even = 10 = 1
numbers from both boxes 50 5

Assignment
T.B Page 75

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 1

Online Class – X − 32 06 / 09 / 2021

3. Mathematics Of Chance – Class 5

To view class

Activity 1

In a class there are 40 students. If a student is selected from

this class the probability that the student selected being a boy
3
is 8

a) What is the number of boys in the class?

b) Find the number of girls in the class .

c) After some more boys joined in the class, the probability that
1
the student selected being a boy became 2 .

Find the number of boys newly joined .

Ans)

Given,

Total number of students = 40 3
8
Probability that the student selected being a boy =

a) Number of boys in the class = 3 of 40
8
3
= 8 × 40 = 3 × 5 = 15

b) Number of girls in the class = 40 − 15 = 25

c) Probability of selecting a boy becomes 1 means
2

number of boys and number of girls are equal .

Since the number of girls in the class = 25 ,

the number of boys in the class = 25

∴ Number of newly joined boys = 25 − 15 = 10

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 2

Activity 2
One is asked to say a two digit number
a)How many two digit numbers are there ?
b)What is the probability that the number said have same

digits?
c)What is the probability that the number said is a perfect

square ?
Ans )
a) Total number of two digit numbers = 99 − 9

= 90

b) Two digit numbers having same digits are
11, 22, 33, 44, 55, 66, 77, 88, 99 (9 numbers)

∴ Probability that the number said have same digits = 9
90
1
= 10

c) Perfect squares having two digits are

16, 25, 36, 49, 64, 81 (6 numbers)

∴ Probability that the number said is a perfect square = 6
90
1
= 15

Activity 3
Two dice with the numbers 1 to 6 are written on their faces are
rolled
a)What is the probability that the numbers so got are same?
b)What is the probability that the first number is less than

second ?
c)What is the probability that their sum is 7 ?

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 3

Ans) 5
4 Pairs with
3 first number

less than the

2 second
1

Sum of the Pairs with
numbers in the same numbers
pair is 7

Total number of pairs = 6 × 6 = 36

a) Number of pairs with same numbers = 6 6 1
36 6
Probability that the numbers are same = =

b) Number of pairs with first number =1+2+3+4+5
less than the second number = 15

∴ Probability of getting pairs with the first = 15 = 5
number is less than the second number 36 12

c) Number of pairs with sum of the numbers 7 = 6

∴ Probability of getting a pair with their sum 7 = 6 = 1
36 6

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 1

Online Class – X − 33 08 / 09 / 2021

3. Mathematics Of Chance – Class 6

To view class

Activity 1

Two coins are tossed together

a) Find the probability that bot the coins show head .

b) Find the probability that both the coins show tail.

c) Find the probability of getting one head and one tail .

Ans)

When two coins are tossed , possible outcomes are

Coin 1 Coin 2

(H,H) (H,T) H H

(T,H) (T,T) T

T

∴ Total number of pairs = 4

a) Pairs with both coins show head − ( H , H ) 1
4
∴ Probability that both the coins show head =

b) Pairs with both the coins show tail − ( T , T )
1
∴ Probability that both the coins show tail = 4

c) Pairs with one head and one tail − ( H , T ) , ( T , H )

Number of pairs = 2 2 1
4 2
∴ Probability of getting one head and one tail = =

Activity 2
There are 30 mangoes in a basket 12 of which are unripe.
Another basket contains 25 mangoes ,with 15 unripe. If we
take one mango from each basket, what is the probability of
both being ripe?

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 2

Ans) Basket 1 Basket 2

12 unripe 15 unripe
18 ripe 10 ripe

30 Mangoes 25 Mangoes

Total number of mangoes in the first basket = 30
Number of mangoes which are unripe =12
Number of mangoes which are ripe = 30 −12 = 18

Total number of mangoes in the second basket = 25

Number of mangoes which are unripe =15

Number of mangoes which are ripe = 25 −15 = 10

Number of pairs with both the mangoes are ripe = 18 × 10

= 180

Total number of pairs = 30 × 25 =750 180 6
750 25
∴ Probability that both the mangoes are ripe = =

Activity 3

When two coins are tossed together, find the probability of

getting at least one head?

Ans) When two coins are tossed , possible outcomes are

Coin 1 Coin 2

(H,H) (H,T) HH

(T,H) (T,T)

TT

∴ Total number of pairs = 4

Pairs having at least one head − ( H , H ), ( H , T ) , ( T , H )

So number of pairs having at least one head = 3
3
∴ Probability of getting at least one head = 4

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 3

Activity 4
In class 10A, there are 30 boys and 20 girls.
In 10B, there are 15 boys and 25 girls.
One student is to be selected from each class.

i) What is the probability of both being girls?
ii) What is the probability of both being boys?
iii) What is the probability of one boy and one girl?
iv) What is the probability of at least one boy?
Ans)

10 A 10 B
30 Boys 15 Boys
20 Girls 25 Girls

Total number of Total number of = 15+25
students in 10 A = 30+20 students in 10 B = 40

= 50

∴ Total no. of possible pairs = 50 × 40 = 2000

i) Number of pairs in which both are girls = 20 × 25

500 = 500
2000
∴ Probability of both being girls =

= 1
4

ii) Number of pairs in which both are boys = 30 × 15
= 450

∴ Probability of both being boys = 450
2000
9
= 40

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Mathematics Of Chance 4

Number of pairs in which = (30 × 25) + (20 ×15)
iii) one boy and one girl = 750 + 300
= 1050

∴ Probability of one boy and one girl = 1050
2000
21
= 40

iv) Number of pairs in = Total no. of − Number of pairs in
which at least one boy possible pairs which both are girls

= 2000 − 500
= 1500

∴ Probability of at least one boy = 1500
2000
3
= 4

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi