Mathematics

A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?

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Out of the two friends, A’s birthday can be any day of the year. Now, B’s birthday can also be any day of 365 days in the year.

We assume that these 365 outcomes are equally likely.

So,

(i) If A’s birthday is different from B’s, the number of favourable outcomes for his birthday is 365 - 1 = 364

P(A’s birthday is different from B’s birthday) = $\dfrac{\text{364}}{\text{365}}$ .

Hence, the probability that both friends will have different birthdays is $\dfrac{364}{365}$

(ii) P(A and B have the same birthday) = 1 - P (both have different birthdays)

= 1 - $\dfrac{364}{365}$

= $\dfrac{365 - 364}{365} = \dfrac{1}{365}$ .

Hence, the probability that both friends will have same birthday is $\dfrac{1}{365}$ .

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