Solution: 2017 Winter Final - 12

Author: Michiel Smid

Question

Let $X = \{1,2,3,\dots,10\}$. We choose, uniformly at random, a subset $Y$ of $X$, where $Y$ has size 5. Define the events

What is $\Pr(A|B)$?

(a)

4/9

(b)

5/9

(c)

5/8

(d)

4/8

Let S be all possible subsets of size 5: $ |S| = \binom{10}{5} $
Let's determine A
First, pick 1: 1
Then we pick 4 more numbers from the remaining 9: $ \binom{9}{4} $
$ |A| = \binom{9}{4} $
$ Pr(A) = \frac{ \binom{9}{4} }{ \binom{10}{5} } = \frac{9!}{5!4!} div \frac{10!}{5!5!} = \frac{9!}{5!4!} \cdot \frac{5!5!}{10!} = \frac{1}{4!} \cdot \frac{5!}{10} = \frac{1}{1} \cdot \frac{5}{10} = \frac{1}{2} $
Let's determine B
First, pick 7: 1
Then we pick 4 more numbers from the remaining 9: $ \binom{9}{4} $
$ |B| = \binom{9}{4} $
$ Pr(B) = \frac{ \binom{9}{4} }{ \binom{10}{5} } = \frac{9!}{5!4!} div \frac{10!}{5!5!} = \frac{9!}{5!4!} \cdot \frac{5!5!}{10!} = \frac{1}{4!} \cdot \frac{5!}{10} = \frac{1}{1} \cdot \frac{5}{10} = \frac{1}{2} $
Let's determine $ A \cap B $
First, pick 1: 1
Second, pick 7: 1
Then we pick 3 more numbers from the remaining 8: $ \binom{8}{3} $
$ |A \cap B| = \binom{8}{3} $
$ Pr(A \cap B) = \frac{ \binom{8}{3} }{ \binom{10}{5} } = \frac{8!}{5!3!} div \frac{10!}{5!5!} = \frac{8!}{5!3!} \cdot \frac{5!5!}{10!} = \frac{8!}{3!} \cdot \frac{5!}{10!} = \frac{1}{3!} \cdot \frac{5!}{10 \cdot 9} = \frac{1}{1} \cdot \frac{5 \cdot 4}{10 \cdot 9} = \frac{5 \cdot 2}{5 \cdot 9} = \frac{2}{9} $

$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } $

$ Pr(A|B) = \frac{ \frac{2}{9} }{ \frac{1}{2} } $

$ Pr(A|B) = \frac{2}{9} \cdot \frac{2}{1} $

$ Pr(A|B) = \frac{4}{9} $