Solution: 2014 Winter Midterm - 16
Author: Michiel SmidQuestion
Consider three people, each one having a uniformly random birthday (out of 365 days; we ignore leap
years). What is the probability that at least two of them have the same birthday?
(a)
$1 - \left. {3 \choose 2} \middle/ 365^3 \right.$
(b)
$1 - \frac{365^2}{364 \cdot 363}$
(c)
$1 - \frac{364 \cdot 363}{365^2}$
(d)
$1 - \left. \bigl\{ {3\choose 2} + {3 \choose 3} \bigr\} \middle/ 365^3 \right.$
Solution
A = All 3 people have different birthdays
First person can have any birthday = 365
Second person can have any birthday except the first person’s = 364
Third person can have any birthday except the first two people’s = 363
$|A| = 365 \cdot 364 \cdot 363$
$ Pr(A) = \frac{365 \cdot 364 \cdot 363}{365^3}$
$ Pr(A) = \frac{364 \cdot 363}{365^2}$
$1 - Pr(A)$ is the probability that at least two of them have the same birthday
$= 1 - \frac{364 \cdot 363}{365^2}$