Few questions from the Class 10 Maths NCERT Exemplar Exercise 13.3 are given below:

Q. Two dice are thrown at the same time. Find the probability of getting

(i) same number on both dice.

(ii) different number on both dice.

Sol.

As the two dice are thrown at the same time,                                

So, total number of possible outcomes = 6² = 36

(i) In case of same numbers on both dice the possible outcomes are

      (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

∴ Number of possible outcomes = 6

So, required probability = 6/36 = 1/6

(ii) In case of different numbers on both dice the possible outcomes are

      = 36 – Number of possible outcomes for same number on both dice

      = 36 – 6 = 30

∴ Required probability = 30/36 = 5/6

Q. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is

(i) 7?                           

(ii) a prime number ?  

(iii) 1?

Sol.

As the two dice are thrown at the same time,                                

So, total number of possible outcomes = 6² = 36

(i) Sum of the numbers appearing on the dice is 7.

Here, the possible outcomes are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6,1),

Thus, the number of possible outcomes = 6

∴  Required probability = 6/36 = 1/6.

(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

Here, the possible ways are (1, 1), (1, 2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3,4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).

Thus, number of possible outcomes = 15

\ Required probability = 15/36 = 5/12.

(iii) Sum of the numbers appearing on the dice is 1.

It is an impossible event, so its probability is zero.

Q. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

(i) 6                            

(ii) 12                         

(iii) 7

Sol.

As the two dice are thrown at the same time,                                

So, total number of possible outcomes = 6² = 36

(i) Product of the numbers on the top of the dice is 6.

Here, the possible ways are (1, 6), (2, 3), (3, 2) and (6, 1).

Thus, the number of possible outcomes = 4

∴ Required probability = 4/36 = 1/9.

(ii) Product of the numbers on the top of the dice is 12.

Here, the possible ways are (2, 6), (3, 4), (4, 3) and (6, 2).

Thus, the number of possible outcomes = 4

∴ Required probability = 4/36 = 1/9.

(iii) Product of the numbers on the top of the dice is 7.

It is an impossible event.

So, its probability is zero.

Q. Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Sol.

As the two dice are thrown at the same time,

Number of total outcomes = 36

For the product of numbers appearing on them to be less than 9, the possible outcomes are

(1, 6), (1,5) (1,4), (1,3), (1,2), (1, 1),(2, 2), (2, 3), (2, 4), (3,2), (4, 2), (4, 1), (3,1), (5,1), (6, 1) and (2,1).

Thus, the number of possible outcomes = 16

∴ Required probability = 16/36 = 4/9