Solution: 2017 Fall Midterm - 17
Author: Michiel SmidQuestion
After having proctored the midterm, Alexa, Farah, May, and Shelly decide to go trick-or-treating.
For any house that the ladies visit, if they do not receive candy, they throw rotten eggs at the
house.
Let $m \geq 7$ and $n \geq 7$ be integers. There are $m$ houses whose owners hand out candy and there are $n$ houses whose owners do not hand out candy.
The ladies choose a uniformly random subset of 7 houses and visit these 7 houses. Define the event
Let $m \geq 7$ and $n \geq 7$ be integers. There are $m$ houses whose owners hand out candy and there are $n$ houses whose owners do not hand out candy.
The ladies choose a uniformly random subset of 7 houses and visit these 7 houses. Define the event
- A = "the ladies throw rotten eggs at exactly 2 of the 7 chosen houses".
(a)
$\frac{{m \choose 5} + {n \choose 2}}{{m+n \choose 7}}$
(b)
$\frac{{7 \choose 2}}{{m+n \choose 7}}$
(c)
$\frac{{7 \choose 2}}{{m \choose 5} \cdot {n \choose 2}}$
(d)
$\frac{{m \choose 5} \cdot {n \choose 2}}{{m+n \choose 7}}$
Solution
If they throw rotten eggs at exactly 2 of the 7 chosen houses, that means they receive candy from 5 of the 7 chosen houses.
We choose 5 houses from the $ m $ houses that hand out candy: $ \binom{m}{5} $
We choose 2 houses from the $ n $ houses that do not hand out candy: $ \binom{n}{2} $
Thus, $ |A| = \binom{m}{5} \cdot \binom{n}{2} $
There are $ \binom{m+n}{7} $ ways to choose a 7-element subset of the $ m+n $ houses.
Thus, $ Pr(A) = \frac{|A|}{\binom{m+n}{7}} $
$ Pr(A) = \frac{\binom{m}{5} \cdot \binom{n}{2}}{\binom{m+n}{7}} $