What is the probability that in a class of $25$ students:
- at least two are born in the same month;
- at least three are born in the same month?
It's obvious that the first probability $P(A)$ is $1$ because there are $12$ months in a year, so there must be two students that are born in the same month. I am not sure that my teacher is ever going to be okay with this solution. I am expected to solve the problem using $Pr(A)=\dfrac{m}{n}$, where $m$ is the number of favorable outcomes, and $n$ is the number of all outcomes. What are the values of $m$ and $n$ in this case? It must be $m=n=12$, but I am not sure I get it. Can someone explain to me why is this?
Best Answer
I think your teacher will accept that. $n = 12^{25}$ is then number of ways in which $25$ kids can be born in one of $12$ months.
$k= 0$ which is the number of ways there are at most $2$ that are born on the same month. And $m = 12^{25}-k = 12^{25}$ are then ways in which at least $3$ are born on the same month.
The question is: how do we calculate that $k=0$. Cite "the pigeon hole principal":
If at most $2$ are born each of the the twelve months then at most $2*12$ are born in twelve of the twelve months. This accounts for at most $24$ kids. There is at least one more kid but no more months available.
SO this can not be done.
That is an acceptable mathematical argument.
So $P($at least 3$) = 1- P($ at most 2$) = 1-\frac {0}{12^{25}} = 1-0 = 1$.