# ICSE Class 8 Maths Selina Solutions for Chapter 23 Probability

### CHAPTER 23-PROBABILITY

Question 1

A die is thrown, find the probability of getting:

(i) A prime number

Solution:-

A die has six numbers: 1, 2, 3, 4, 5, 6

No. of possible outcomes =6

We know that

Number of favorable outcomes = a prime number = 1, 3, 5 which are 3 in numbers (Formula)

Here

$P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{3}{6}=\frac{1}{2}$

(ii) A number greater than 4

Solution:-

We know that

No. of favorable outcome = Greater than four i.e. two number 5 and 6

Here

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{2}{6}=\frac{1}{3}$

(iii) A number not greater than 4.

Solution:-

We know that

Number of favorable outcome = not greater than 4 or numbers will be 1,2,3,4 which are 4 in numbers

Here

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{4}{6}=\frac{2}{3}$

Question 2.

A coin is tossed. What is the probability of getting

(i) A tail?

Solution:-

On tossing a coin once,

No. of possible outcome =2

(i) Favorable outcome getting a tail =1

No. of favorable outcome =2

We know that

$P(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{1}{2}$

Solution:-

Favorable outcome getting a head  =1

No. of possible outcome =2

We know that

$P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{1}{2}$

Question 3.

A coin is tossed twice. Find the probability of getting:

Solution:-

Possible number of favorable outcomes =2  (i.e. TH and HT)

Total number of possible outcomes =4

We know that

$∴ \mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}=\frac{2}{4}=\frac{1}{2}$

(ii) Exactly one tail

Solution:-

Exactly one tail

Possible number of favorable outcomes =2 (i.e. TH and HT)

Total number of possible outcomes =4

We know that

$P(E)=\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}=\frac{2}{4}=\frac{1}{2}$

(iii) Two tails

Solution:-

Possible number of favorable outcomes =1  (i.e. TT)

Total number of possible outcomes =4

We know that

$P(E)=\frac{\text {Number of favourable outcomes}}{\text {Total number of possible outcomes}}=\frac{1}{4}$

Solution:-

Possible number of favorable outcomes =1 (i.e. HH)

Total number of possible outcomes = 4

So we get

$P(E)=\frac{1}{4}$

Question 4.

A letter is chosen from the word ‘PENClL’ what is the probability that the letter chosen is a consonant?

Solution:-

We know that

Total no. of letters in the word ‘PENCIL’ =6

Total Number of Consonant = ‘PNCL’ i.e, 4

Here

$P(E)=\frac{\text { Total No.of consonants }}{\text { Total No.of Letters in the word PENCIL }}=\frac{4}{6}=\frac{2}{3}$

Question 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

Solution:-

Total number of possible outcomes =3

$\mathrm{P}(\mathrm{E})=\frac{1}{3}$

(ii) Not a red ball

Solution:-

No. of favorable outcomes

$\mathrm{P}(\mathrm{E})=\frac{2}{3}$

(iii) A white ball.

Solution:-

No. of favorable outcomes =0

$\mathrm{P}(\mathrm{E})=\frac{0}{3}=0$

Question 6.

In a single throw of a die, find the probability of getting a number

(i) Greater than 2

Solution:-

A die has six numbers =1, 2, 3,4,5,6

No. of possible outcomes =6

$\mathrm{P}(\mathrm{E})=\frac{4}{6}=\frac{2}{3}$

(ii) less than or equal to 2

Solution:-

Number of favorable outcomes =1, 2

$\mathrm{P}(\mathrm{E})=\frac{2}{6}=\frac{1}{3}$

(iii) Not greater than 2.

Solution:-

Number of favorable outcomes =1, 2

$\mathrm{P}(\mathrm{E})=\frac{2}{6}=\frac{1}{3}$

Question 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) Number of possible outcomes of one black ball =10 and number of favorable outcome of one black ball =5

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourableoutcome }}{\text { Number of all possible outcome }}=\frac{5}{10}=\frac{1}{2}$

(ii) Number of possible outcome of one red ball =10 and number of favorable outcome of one red ball =2

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourableoutcome }}{\text { Number of all possible outcome }}=\frac{2}{10}=\frac{1}{5}$

(iii) Number of possible outcome of one White ball =10 and number of favorable outcome =3

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourableoutcome }}{\text { Number of all possible outcome }}=\frac{3}{10}$

(iv) Number of possible outcome =10

Number of favorable outcome  = 3+5=8

Not a red ball

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourableoutcome }}{\text { Number of all possible outcome }}=\frac{8}{10}=\frac{4}{5}$

(v) Number of possible outcome  =10

Number of favorable outcome not a red ball=3+2=5

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourableoutcome }}{\text { Number of all possible outcome }}=\frac{5}{10}=\frac{1}{2}$

Question 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 favorable outcome = an even number i.e.  2, 4, 6 which are 3 in numbers

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{3}{6}=\frac{1}{2}$

(ii)&(iii) Number of favourable outcome = not an even number i.e. odd numbers

: 1,3,5 which are 3 in numbers

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{3}{6}=\frac{1}{2}$

Question 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 favorable outcome =0

(∵8 is not possible)

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{0}{6}=0$

(ii) Number greater than 8 will be 0

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{0}{6}=0$

(iii) Number less than 8 will be 1,2,3,4, 5, 6

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{6}{6}=1$

Question 10.

Which of the following cannot be the probability of an event?

(i) $\frac{2}{7}$

(ii) 3.8

(iii) 37%

(iv) -0.8

(v) 0.8

(vi) $\frac{-2}{5}$

(vii) $\frac{7}{8}$

Solution:-

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

(vi) $-\frac{2}{5}$, because probability of an even can never be less than 1.

Question 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

Probability is zero (0) = or P(E)=0

(ii) a black ball

Number of favorable outcome =1

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{1}{6}$

Question 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) Favorable outcome= i.e. (H,H,H)

P(of getting all heads ) $=\frac{1}{8}$

(ii) Favorable outcomes =3(H, H, T),(H,T, H),(T,H,H)

$P(E)=\frac{3}{8}$

(iii) Favorable outcomes =3(H,T,H),(T,T,H),(H,T,T)

$P(E)=\frac{3}{8}$

(iv) Favorable outcomes =1 i.e. (T,T,T)

$P(E)=\frac{1}{8}$

Question 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

Here

Number of pages whose sum of its page is 9=10

i.e. 9,18,27,36,45,54,63,72,81,90

$\mathrm{P}(\mathrm{E})=\frac{10}{92}=\frac{5}{46}$

Question 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 Tail or H.T

Two coins are tossed

Here

Number of coins $=2 \times 2=4$ which are HH, HT, TH, TT

(i) At least one head, then Number of outcomes =3

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{3}{4}$

(ii) When both head or both tails, then

Number of outcomes =2

We know that

$\mathrm{P}(\mathrm{E})=\frac{\text { Number of favourable outcome }}{\text { Number of all possible outcome }}=\frac{2}{4}=\frac{1}{2}$

Question 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) Favorable outcomes =5 i.e. 2, 4, 6, 8, 10

$P(E)=\frac{5}{10}=\frac{1}{2}$

(ii) Favorable outcomes =3 i.e. 3, 6, 9

$P(E)=\frac{3}{10}$

(iii) Favorable outcomes =1 i.e. 6

$P(E)=\frac{1}{10}$

(iv) Favorable outcomes =7

i.e. 2, 3, 4, 6, 8, 9, 10

$\mathrm{P}(\mathrm{E})=\frac{7}{10}$