# Selina Concise Solutions for Chapter 23 Probability Class 8 ICSE Mathematics

## ### Exercise 23

1. A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Solution
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

(i) A prime number
Number of favourable outcomes = a prime number = 1, 3, 5 which are in numbers
P(E) = (Numbers of favourable outcome)/(Numbers of all outcome)
= 3/6 = ½

(ii) Numbers of favourable outcome = greater than four i.e two number 5 and 6
∴ P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)
= 2/6 = 1/3

(iii) Number of favourable outcome = not graeter than 4 or numbers will be 1, 2, 3, 4 which are 4 in numbers
∴ P(E) = (Numbers of favourable outcome)/(Number of all outcome)
= 4/6
= 2/3

2. A coin is tossed. What is the probability of getting:
(i) a tail?
Solution
On tossing a coin once,
Number of possible outcome = 2

(i) Favourable outcome getting a tail = 1
⇒ number of favorable outcome = 2
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= ½

Similarly, favourable outcome getting a head = 1
But number of possible outcome = 2
∴  P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= ½

3. A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Solution
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½

(ii) Exactly one tail
Possible number of favourable outcomes = 2
(i.e. TH and HT)
Total number of possible outcomes = 4
P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= 2/4
= ½

(iii) Two tails
Possible number of favourable outcomes = 1
(i.e., TT)
Total number of possible outcome s = 4
∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)
= ¼

Possible number of favourable outcomes = 1
(i.e. HH)
Total number of possible outcomes = 4
∴ P(E) = ¼

4. A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Solution
Total no. of letters in the word ‘PENCIL = 6
Total Number of Consonant = ‘PNCL’  i.e. 4
P(E) = (Total No. of consonants)/(Total No. of letters in the word PENCIL)
= 4/6
= 2/3

5. A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Solution
(i) Total number of possible outcomes = 3
∴ P(E) = 1/3
(ii) Not a red ball
Number of favourable outcomes = Green ball + Black Ball
= 1 + 1 + 2
∴ P(E) = 2/3
(iii) A white ball
Number of favourable outcome = 0
∴ P(E) = 0/3
= 0

6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Solution
(i) A die has six numbers = 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6
∴ P(E) = 4/6 = 2/3
(ii) Less than or equal to 2
Number of favourable outcome = 1, 2
∴ P(E) = 2/6
= 1/3
(iii) Not greater than 2
Number of favourable outcomes = 1, 2
∴ P(E) = 2/6
= 1/3

7. A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.
A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution
In a bag, 3 balls are white
2 balls are red
5 balls are black
Total number of balls = 3 + 2 + 5 = 10

(i) Numbers of possible outcome of one black ball = 10
And number of favourable outcome of one black ball = 5
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 5/10
= 1/2

(ii) Number of possible outcome of one red ball = 10
And number of favourable outcome = 2
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 2/10
= 1/5

(iii) Number of possible outcome of white ball = 10
And number of favourable outcome = 3
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 3/10

(iv) Number of possible outcome = 10
Number of favourable outcome = 3 + 5 = 8
Not a red ball
∴ P(E)  = (Number of favourable outcome)/(Number of all possible outcome)
= 8/10
= 4/5

(v) Number of possible outcomes = 10
Number of favourable outcome not a black ball = 3 + 2 = 5
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome) = 5/10
= 1/2

8. In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Solution
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcome = 6

(i) Number of favourable outcome = an even number i.e. 2, 4, 6 which are 3 in numbers
∴  P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)
= 3/6
= 1/2

(ii) & (iii) Number of favourable outcome = not an even number i.e. odd numbers : 1, 3, 5 which are 3 in numbers
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 3/6
= 1/2

9. In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Solution
On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6
(i) Number of favourable outcome = 0
(∵ 8 is not possible)
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 0/6 = 0

(ii) Number greater than 8 will be 0
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 0/6
= 0

(iii) Number less than 8 wil be 1, 2, 3, 4, 5, 6
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 6/6
= 1

10. Which of the following can not be the probability of an event?
(i) 2/7
(ii) 3.8
(iii) 37%
(iv) - 0.8
(v) 0.8
(vi) -2/5
(vii) 7/8
Solution
(i) The probability of an event cannot be
(ii) 3.8 i.e. the probability of an even cannot exceed 1.
(iv) i.e. -0.8 and
(vi) -2/5, This is because probability of an even can never be less than 1.

11. A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Solution
∵ There are 6 black balls in a bag
∴ number of possible outcome = 6

(i) A white ball
As there is no white ball in the bag
∴ Its probability is zero (0) = or P(E) = 0

(ii) a black ball
∴ Number of favourable outcome = 1
∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 1/6

12. Three identical coins are tossed together. What is the probability of obtaining:
Solution
Total outcomes = 8
i.e., (H, H, H), (H, H, T), (H, T, H), (T, T, T), (T, H, H), (T, T, H), (H, T, T), (T, H, T)
(i) Favourable outcome = i.e., (H, H, H)
∴ P (of getting all heads) = 1/8

(ii) Favourable outcome = 3 (H, H, T), (H, T, H), (T, H, H)
∴ P(E) = 3/8

(iii) Favourable outcome = 3 (H, T, H), (T, T, H), (H, T, T)
∴ P(E) = 3/8

(iv) Favourable outcomes = 1  i.e., (T, T, T)
∴ P(E) = 1/8

13. A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Solution
Number of pages of the book = 92
Which are from 1 to 92
Number of possible outcomes = 92
∴ Number of pages whose sum of its page is 9 = 10
i.e. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90
∴ P(E) = 10/92
= 5/46

14. Two coins are tossed together. What is the probability of getting:
(ii) both heads or both tails.
Solution
∵ A coins has two faces Head and Tailor H, T
∴ Two coins are tossed
∴ Number of coins = 2 x 2 = 4
which are HH, HT, TH, TT

(i) At least one head, then
Number of outcomes = 3
∴  P(E) = (Number of favourable outcome)/(Number of all possible outcome)
= 3/4

(ii) When both head or both tails, then
Number of outcomes = 2
∴ P(E) = (Numberof favourable outcome)/(Number of all possible outcome)
= 2/4
= 1/2

15. From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2 (ii) 3
(iii) 2 and 3 (iv) 2 or 3
Solution
Total outcomes = 10
i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

(i) Favourable outcomes = 5 i.e., 2, 4, 6, 8, 10
P(E) = 5/10 = 1/2

(ii) Favourable outcomes = 3 i.e., 3, 6, 9
P(E) = 3/10

(iii) Favourable outcomes = 1 i.e., 6
P(E) = 1/10

(iv) Favourable outcome = 7
i.e., 2, 3, 4, 6, 8, 9, 10
P(E) = 7/10

16. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Solution
Total outcomes = 36 i.e.
(1, 1) (1, 2) (1, 3) (1, 4) (1,5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3 ,4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

(i) Favourable outcomes = 0
P(E) = 0/36 = 0

(ii) Favourable outcomes = i.e., (6,6)
P(E) = 1/36

(iii) Favourable outcomes = 35  [Except  (6, 6)]
P(E) = 35/36

(iv) Favourable  outcome = 36
P(E) = 36/36
= 1

17. A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Solution
Total outcomes = 6
i.e., 1, 2, 3, 4, 5 and 6

(i) Favourable outcomes = 3 i.e., 2, 3, 5
P(E) = 3/6
= 1/2

(ii) Favourable outcomes = 3 i.e., 4, 5, 6
P(E) = 3/6
= 1/2

(iii) Favourable outcomes = 4 i.e.,  1, 2, 4, 6
P(E) = 4/6
= 2/3

(iv) Favourable outcomes = 5
i.e., 1, 2, 3, 4 and 5
P(E) = 5/6

(v) Favourable outcomes = 0
P(E) = 0/6 = 0

18. Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
Solution
Total outcomes = 4
i.e., HH, HT, TT, TH

(i) Favourable outcomes = 2 i.e., HT and TH
P(E) = 2/4
= 1/2

(ii) Favourable outcomes = 3
i.e., HH, HT and TH
P(E) = 3/4

(iii) Favourable outcomes = 1
i.e., TT
P(E) = 1/4

(iv) Favourable outcomes = 3
i.e., HH, HT and TH
P(E) = 3/4