Question

What is the probability that in a group of $N$ people, at least two of them will have the same birthday?

• A. $\frac{365!/(365-N)!}{{365}^N}$
• B. $1-\frac{365!/(365-N)!}{{365}^N}$
• C. $1-\frac{365!/\left(N\right)!}{{365}^N}$
• D. $\frac{365!/\left(N\right)!}{{365}^N}$

Answer 1

The correct option is B $1-\frac{365!/(365-N)!}{{365}^N}$

We first find the probability that no two persons have the same birthday and then subtract the result from 1.

Excluding leap years,there are $365$ different birthdays possible. Any person might have any one of the $365$ days of the year as a birthday.

A second person may likewise have any one of $365$ birthday : and so on.

Hence the total number of ways is given by $n={365}^N.$

And the number of possible ways for none of the $N$ birthdays to coincide is $m=\mathrm{365.364......}(365-\overline{N-1})$

$=\frac{365!}{365-N!}$

$\therefore$ The probability that no two birthdays coincide is $\frac{m}{n}=\frac{365!/(365-N)!}{{365}^N}$

Hence the required probability $=1-\frac{m}{n}=1-\frac{365!/(365-N)!}{{365}^N}$