```Question 331607
10 students in a classroom -- 3 were born on the same day
What is the probability of this happening? 1 in ????
<pre><b>
First we find the probability of the complement event,
that they all have different birthdays.  Then we subtract
from 1.

Number the students 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

There are 365 birthdays student 1 could have.

Then that leaves 364 birthdays student 2 could have and not have the same birthday as student 1.

Then that leaves 363 birthdays student 3 could have and not have the same birthday as either student 1 or student 2.

etc. etc. until

Then that leaves 356 birthdays student 10 could have and not have the same birthday as either student 1 or student 2.

So the numerator of the probability of the complement event is

365*364*363*362*361*360*359*358*357*356

The denominator of the complement event is found this way:

There are 365 birthdays student 1 could have.
There are 365 birthdays student 2 could have.
There are 365 birthdays student 3 could have.

etc. etc. until

There are 365 birthdays student 10 could have.

So the denominator of the probability of the complement event is {{{365^10}}}.

So the probability of the complement event is

{{{(365P10)/(365^10)}}}

That's approximately 0.8830518223

So the desired probability is 1 - 0.8830518223 or .1169481777.