Solution: 2015 Fall Final - 16

Author: Michiel Smid

Question

You are given that:

The course COMP 9999 runs over a period of one year, starting on January 1 and ending on December 31. There is one lecture every day; thus, the total number of lectures is 365.
Dania and Nick take this course. Dania's birthday is on November 19. Nick's birthday is on December 3.
Professor G. Ruesome teaches the course. Professor Ruesome decides to have 20 quizzes during the year. For this, he chooses a uniformly random subset of 20 days; the quizzes will be on the 20 chosen days. (It is possible that there is a quiz on January 1.)

Determine the conditional probability $\Pr(B|C)$, where $B$ and $C$ are the events

(a)

5/32

(b)

5/31

(c)

4/31

(d)

4/32

Let's determine C
We choose 5 of the 31 days in December: $ \binom{31}{5} $
The remaining 15 quizzes can be on any of the 334 days in the year: $ \binom{334}{15} $
$ |C| = \binom{31}{5} \cdot \binom{334}{15} $
$ Pr(C) = \frac{\binom{31}{5} \cdot \binom{334}{15}}{\binom{365}{20}} $
Let's determine $ B \cap C $
1 of the days in December is Nick's birthday, so that's fixed: 1
We choose 4 of the remaining 30 days in December: $ \binom{30}{4} $
The remaining 15 quizzes can be on any of the 334 days in the year: $ \binom{334}{15} $
$ |B \cap C| = \binom{30}{4} \cdot \binom{334}{15} $
$ Pr(B \cap C) = \frac{\binom{30}{4} \cdot \binom{334}{15}}{\binom{365}{20}} $
$ Pr(B|C) = \frac{Pr(B \cap C)}{Pr(C)} $
$ Pr(B|C) = \frac{ \frac{\binom{30}{4} \cdot \binom{334}{15}}{\binom{365}{20}}}{ \frac{\binom{31}{5} \cdot \binom{334}{15}}{\binom{365}{20}}} $
$ Pr(B|C) = \frac{\binom{30}{4}}{\binom{31}{5}} $
$ Pr(B|C) = \frac{ \frac{30!}{4! \cdot 26!}}{ \frac{31!}{5! \cdot 26!}} $
$ Pr(B|C) = \frac{30! \cdot 26! \cdot 5!}{4! \cdot 26! \cdot 31!} $
$ Pr(B|C) = \frac{30! \cdot 5!}{4! \cdot 31!} $
$ Pr(B|C) = \frac{30! \cdot 5 \cdot 4!}{4! \cdot 31 \cdot 30!} $
$ Pr(B|C) = \frac{5}{31} $