# 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

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