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

∴ Number of possible outcomes =6

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

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

 (ii) A number greater than 4 

Solution:-

Number of favorable outcome = Greater than four i.e. two number 5 and 6

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

(iii) A number not greater than 4. 

Solution:-

(iii) Number of favorable outcome = not greater 

Than 4 or numbers will be 1,2,3,4 which are 4 in numbers 

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

Question 2. 

 A coin is tossed. What is the probability of getting

 (i) A tail?

Solution:-

 On tossing a coin once, Number of possible outcome =2

(i) Favorable outcome getting a tail =1

 Number of favorable outcome =2

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

(ii) a head? 

Solution:-

A head

 Similarly, favorable outcome getting a head  =1

 But number of possible outcome =2

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

Question 3. 

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

 (i) Exactly one head

Solution:-

Exactly one head 

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

Total number of possible outcomes =4

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

(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

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

(iii) Two tails

Solution:-

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

Total number of possible outcomes =4

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

(iv) Two heads 

Solution:-

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

Total number of possible outcomes = 4

\(∴ P(E)=\frac{1}{4}\)

Question 4. 

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

Solution:-

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

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

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

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:-

Number of favorable outcomes 

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

 (iii) A white ball. 

Solution:-

Number 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

∴ Number 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

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

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

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

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

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

(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

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

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

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

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)

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

(ii) Number greater than 8 will be 0

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

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

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

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 and 

(vi) \(-\frac{2}{5}\), This is 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

∴ Its probability is zero (0) = or P(E)=0

(ii) a black ball

∴ Number of favorable outcome =1

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

Question 12.

Three identical coins are tossed together. What is the probability of obtaining?

(i) All heads?

(ii) Exactly two heads?

(iii) Exactly one head?

(iv) No head?

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

∴ 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:

(i) at least one head

(ii) both heads or both tails.

Solution:-

∵ A coins has two faces Head and Tail or H.T

∴ Two coins are tossed

∴ 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

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

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

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

Leave a Comment

Your email address will not be published. Required fields are marked *