Q. 2.46 - A First Course in Probability

Question

How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least $\frac{1}{2}?$ Assume that all possible monthly outcomes are equally likely.

Step-by-Step Solution

Answer

$n=5$

Calculate the probability for any$n$.

Step 1 Given Information.

Assume that all possible monthly outcomes are equally likely.

Step 2 Explanation.

If $n$people are in the room:

The outcome space of the experiment is$S\left\{\left(x_1,x_2,x_3,\dots x_n\right):x_i\in \{1,2,3,\dots 12\}\right\}$

where the vector describes their months of birth.

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

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

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

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

Remember proposition $4.1$

$P\left(A^c\right)=1-P\left(A\right)$

From this:

$P\left(\text{ at least two share birth months }\right)=1-P\left(\text{ no two people share birth months }\right)$

And the number of sample events in the event on the right-hand side is $12·11·10·\dots (12-n+1)=\frac{12!}{(12-n)!}$the number of different valued vectors of size$n$.

$P( at least two share birth months )=1-\frac{\frac{12!}{(12-n)!}}{{12}^n}$

$$\begin{gathered} =1-\frac{12·11·10·\dots (12-n+1)}{{12}^n} \\ =1-\frac{11}{12}·\frac{10}{12}·\dots \frac{(12-n+1)}{12} \end{gathered}$$

Step 3 Explanation.

For what $n$is this probability$>\frac{1}{2}$.

$n=3 1-\frac{11}{12}·\frac{10}{12}=1-\frac{110}{144}=\frac{34}{144}\le \frac{1}{2}$

$n=4 1-\frac{11}{12}·\frac{10}{12}·\frac{9}{12}=1-\frac{990}{1728}=\frac{738}{1728}\le \frac{1}{2}$

$n=5 1-\frac{11}{12}·\frac{10}{12}·\frac{9}{12}·\frac{8}{12}=1-\frac{7920}{20736}=\frac{12716}{20736}>\frac{1}{2}$

For $n=5$in the room, there is more than$50\%$ a chance that the two of them share the month of birth.