Question

The possibility that two friends have different birthday non leap year

Answer 1

$Here\: is\: your \: answer\: buddy\:$

$\textbf{GIVEN}$

It is a non leap year. That means the year consist of 365 days.

$\textbf{NOW}$

The birthday of first friend will be on any day of the total 365 days.

So, P(1) = $\frac{365}{365}$

$\textbf{AGAIN}$

The birthday of the 2nd friend will be on any day of the total of 364 days as his birthday date will be on the different day than the first friend.

So, P(2) = $\frac{364}{365}$

So, P = $\frac{365}{365} × \frac{364}{365}$

P = $\frac{364}{365}$

Hope this helps you.
Be Brainly.

Answer 2

Given,

Two friends have different  birthday in non leap year.

We know  that  a  non leap year contains  365 days  while  there are 366 days  in a  leap year.

Now,

If both have same birthday, then it may  be  any  of  one  day from  365 days.

It  means  favourable  case is  one  while  total number of  possible  outcomes  is  365. Hence,

Probability of having  same  birthday = 1/365.

Therefore, probability of  not having  same  birthday = 1 - 1/365

                                                                                       = (365-1)/365

                                                                                       = 364/365