Solution: 2016 Fall Final - 16

Author: Michiel Smid

Question

Let $n \geq 2$ be the number of students who are writing this exam. Each of these students has a uniformly random birthday, which is independent of the birthdays of the other students. We ignore leap years; thus, the year has 365 days. Define the event

A = "at least two students have their birthday on December 14".

What is $\Pr(A)$?

(a)

${\sum_{k=2}^{n}} {n \choose k} \cdot \left( \frac{1}{365} \right)^k$

(b)

$1 - \left( \frac{364}{365} \right)^n - n \cdot \frac{1}{365} \cdot \left( \frac{364}{365} \right)^{n-1}$

(c)

${n \choose 2} \cdot \left( \frac{1}{365} \right)^2 \cdot \left( \frac{364}{365} \right)^{n-2}$

(d)

$1 - {n \choose 2} \cdot \left( \frac{1}{365} \right)^2 \cdot \left( \frac{364}{365} \right)^{n-2}$

Solution

Let's determine $ S $
S is the set of all possible outcomes of the birthdays of the students
$ |S| = 365^n $
$ Pr(S) = \frac{365^n}{365^n} = 1 $
Let's determine $ B $
B is the set of all outcomes where no students have their birthday on December 14
The first student has 364 possible birthdays
The second student has 364 possible birthdays
...
The nth student has 364 possible birthdays
$ |B| = 364^n $
$ Pr(B) = \frac{364^n}{365^n} $
Let's determine $ C $
C is the set of all outcomes where exactly one student has their birthday on December 14
Out of n students, choose 1 of them to have their birthday on December 14: $ \binom{n}{1} = n $
The student with the birthday on December 14 has 1 possible birthday: 1
The other students have 364 possible birthdays: $ 364^{n-1} $
$ |C| = n \cdot 1 \cdot 364^{n-1} $
$ Pr(C) = \frac{n \cdot 1 \cdot 364^{n-1}}{365^n} $

Now, let’s determine $ A $

$ Pr(A) = Pr(S) - Pr(B) - Pr(C) $

$ Pr(A) = 1 - \frac{364^n}{365^n} - \frac{n \cdot 1 \cdot 364^{n-1}}{365^n} $