In a single throw of 2 die, we have total 36(6 × 6) outcomes possible.

Say, n (S) = 36

Where, ‘S’ represents sample space

Let ‘A’ denotes the event of getting a double.

So, A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

P (A) = \( \dfrac{ n (A) }{ n (S)}\)

= \( \dfrac{6}{36}\)

= \( \dfrac{1}{6}\)

And ‘B’ denotes the event of getting a total of 9

So, B = {(3,6), (6,3), (4,5), (5,4)}

P (B) =\( \dfrac{ n (B) }{ n (S)}\)

= \( \dfrac{4}{36}\)

= \( \dfrac{1}{9}\)

We need to find probability of neither the event of getting neither a doublet nor a total of 9.

P (A′ ∩ B′) =?

As, P (A′ ∩ B′) = P (A ∪ B)′ {using De Morgan’s theorem}

P (A′ ∩ B′) = 1 – P (A ∪ B)

By using the definition of P (E or F) under axiomatic approach (also called addition theorem) we know that:

P (E ∪ F) = P (E) + P (F) – P (E ∩ F)

P (A ∪ B) = \( \dfrac{1}{6}\) + \( \dfrac{1}{9}\)+ 0

= \( \dfrac{5}{18}\) {Since, P (A ∩ B) = 0 since nothing is common in set A and B.

So, n (A ∩ B) = 0}

Hence,

P (A′ ∩ B′) = 1 – \( ( \dfrac{5}{18})\)

= \( \dfrac{13}{18}\)

Answered by Aaryan | 1 year agoOne number is chosen from numbers 1 to 100. Find the probability that it is divisible by 4 or 6?

The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75. What is the probability of passing the Hindi examination?

A card is drawn from a deck of 52 cards. Find the probability of getting an ace or a spade card.

A die is thrown twice. What is the probability that at least one of the two throws come up with the number 3?

A natural number is chosen at random from amongst first 500. What is the probability that the number so chosen is divisible by 3 or 5?