Jayanti throws a

Pihu has a better chance of getting number 36

Let, S be the event that contains all possibilities/ combinations occur when Jayanti throws a pair of dice.

There will be total 6 × 6 = 36 outcomes.

∴ n(S) = 36

Let, A be the event such that Jayanti will get the product of the numbers appearing on the dices as 36.

As there is only one possibility (6, 6) for getting product 36.

∴ n(A) = 1

Therefore, the probability of getting product of the numbers appearing on the dice as 36 is

Now, let, S’ be the event that contains all possibilities that Pihu gets after throwing 1 dice.

∴ n(S’) = 6

Let, B be the event such that Pihu will get square of the number appearing on the dice as 36.

As there is only one possibility 6 of which square is 36.

∴ n(B) = 1

Therefore, the probability of getting square of the number appearing on the dice as 36 is

As is greater than , hence, Pihu has a better chance of getting number 36.

OR

(a) (b)

Let, S be the sample space that contains numbers between 70 and 100.

S = {70, 71, 72, 73, …….., 99}

∴ n(S) = 100 – 70 = 30

Now, let A be the set which contains prime numbers between 70 and 100.

∴ A = {71, 73, 79, 83, 89, 97}

∴ n(A) = 6

Therefore, the probability that an integer chosen between 70 and 100 is a prime number is

Now, let B be the set which contains numbers between 70 and 100 which are divisible by 7.

∴ B = {70, 77, 84, 91, 98}

∴ n(B) = 5

Therefore, the probability that an integer chosen between 70 and 100 is divisible by 7 is