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The probability that in a group of n people, at least two of them will have the same date of birth of a non leap year (Assuming a year is having 365 days)

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$\frac{^{365} P_{n}}{{(365)}^{n}}$

$1 - \frac{^{365} P_{n}}{{(365)}^{n}}$

$\frac{1}{{(365)}^{n}}$

$\frac{365 \times 364}{{(365)}^{n}}$

answer is A.

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Detailed Solution

$n (S) = {(365)}^{n}$

Let $\bar{A}$ be the event none of n having some birth date

$n (\bar{E}) =^{365} P_{n}$

$P (\bar{A}) = \frac{^{365} P_{n}}{{(365)}^{n}}$

$P (A) + P (\bar{A}) = 1$

$P (A) = 1 - P (\bar{A})$

$= 1 - \frac{^{365} P_{n}}{{(365)}^{n}}$

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