Solution: 2019 Fall Final - 15

Author: Michiel Smid

Question

A bowl contains 5 blue balls and 4 red balls. We choose 2 balls uniformly at random. Define the events

What is the conditional probability $\Pr(A|B)$?

(a)

$\frac{{4 \choose 2}}{{5 \choose 2} + {4 \choose 2}}$

(b)

$\frac{{5 \choose 2} + {4 \choose 2}}{{4 \choose 2}}$

(c)

$\frac{{4 \choose 2}}{{5 \choose 2} \cdot {4 \choose 2}}$

(d)

$\frac{{4 \choose 2}}{{9 \choose 2}}$

Let's determine $B$
We can choose 2 of the blue balls: $ \binom{5}{2} $
We can also choose 2 of the red balls: $ \binom{4}{2} $
$ |B| = \binom{5}{2} + \binom{4}{2} = 10 + 6 = 16 $
Let's determine $ A \cap B $
We can choose 2 of the red balls: $ \binom{4}{2} $
$ |A \cap B| = \binom{4}{2} = 6 $

Now, let God do the rest

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

$ Pr(A|B) = \frac{ \binom{5}{2} + \binom{4}{2} }{ \binom{4}{2} } $