Question

If there are $12$ strangers in a room, what is the probability that no two of them celebrate their birthday in the same month?

Short answer

The required probability is

$P\left(\text{no two people share their month of birth}\right)=\frac{12!}{{12}^{12}}$.

Step-by-step solution

Given Information 

If $12$people are in the room:

Outcome space of the experiment is $S\left\{\left(x_1,x_2,x_3,\dots x_{12}\right):x_i\in \{1,2,3,\dots 12\}\right\}$ where the vector describes their months of birth.

Explanation

$\left|S\right|={12}^{12}$ because each element of the vector can be chosen in $12$ ways.

As each outcome from the outcome space is equally likely, the Axioms give:

$A\subseteq S\to P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

( $\left|X\right|$ is the number of elements in the set $X$ )

And the number of sample events in the event that no two people share their month of birth is $12·11·10·\dots 1=12!$, the number of different valued vectors in $S$of size $12$.

$P\left(\text{no two people share their month of birth}\right)=\frac{12!}{{12}^{12}}$