Solution: 2016 Fall Midterm - 16
Author: Michiel SmidQuestion
Consider 4 people, each of which has a uniformly random birthday. We ignore leap years; thus, one year
has 365 days. Define the event
- A = "at least 2 of these 4 people have the same birthday".
(a)
$1 - \frac{362 \cdot 363 \cdot 364}{365^3}$
(b)
${4 \choose 2} \cdot \frac{1}{365} + {4 \choose 3} \cdot \frac{1}{365^2} + {4 \choose 4} \cdot \frac{1}{365^3}$
(c)
${4 \choose 2} \cdot \frac{1}{365}$
(d)
$1 - \frac{361 \cdot 362 \cdot 363 \cdot 364}{365^4}$
Solution
We can calculate the probability that none of the 4 people have the same birthday.
B = No one has the same birthday.
The first person has 365 choices.
The second person has 364 choices.
The third person has 363 choices.
The fourth person has 362 choices.
$ |B| = 365 \cdot 364 \cdot 363 \cdot 362 $
$ Pr(A) = 1 - \frac{|B|}{365^4} $
$ Pr(A) = 1 - \frac{365 \cdot 364 \cdot 363 \cdot 362}{365^4} $
$ Pr(A) = 1 - \frac{364 \cdot 363 \cdot 362}{365^3} $