If there are 25 people in a room, what is the
probability that at least two of them share the same birthday.
If your first thought is that it is 25/365 = 0.068, you will be
surprised to learn it is much higher than that. This problem requires
the application of the sections on P(A and B) and conditional
probability.

This problem is best approached by asking what is
the probability that no two people have the same birthday. Once
we know this probability, we can simply subtract it from 1 to
find the probability that two people share a birthday.

If we choose two people at random, what is the probability
that they do not share a birthday? Of the 365 days on which the
second person could have a birthday, 364 of them are different
from the first person's birthday. Therefore the probability is
364/365. Let's define P2 as the probability that the second person
drawn does not share a birthday with the person drawn previously.
P2 is therfore 364/365. Now define P3 as the probability that
the third person drawn does not share a birthday with anyone drawn
previously given that there are
no previous birthday matches. P3 is therefore a conditional probability.
If there are no previous birthday matches, then two of the 365
days have been "used up," leaving 363 non-matching days.
Therefore P3 = 363/365. In like manner, P4 = 362/365, P5 = 361/365,
and so on up to P25 = 341/365.

In order for there to be no matches, the second
person must not match any previous person and
the third person must not match any previous person, and
the fourth person must not match any previous person, etc. Since
P(A and B) = P(A)P(B), all we have to do is multiply P2, P3, P4
...P25 together. The result is 0.431. Therefore the probability
of at least one match is 0.561.