Same Birthdays If 25 people are randomly selected, find the probability that no 2 of them have the same birthday. Ignore leap years.

Twenty-five people are selected at random.

When multiple selections (equal to say n) are made from a sample, the probability of occurrence of an event at difference sequences of selections is given by:

$P\left(A\right)=P\left(Aat{1}^{st}selection\right)\times P\left(Aat{2}^{nd}selection\right)\times .....\times P\left(Aat{n}^{th}selection\right)$

The total number of days in a year is 365.

The total number of people(days) to be selected is 25.

For all birthdays to be unique, the number of favorable options decreases by one at every successive individual.

However, the total number of options remains the same.

For no two people to have the same birthday, the following probability is calculated:

$$\begin{gathered} P\left(alluniquebirthdays\right)=\frac{365}{365}\times \frac{364}{365}\times \frac{363}{365}\times ......\times \frac{341}{365}\left(25\text{selections}\right) \\ =0.4313 \end{gathered}$$

Therefore, the probability that out of 25 selected people, all will have unique birthdays is 0.431.