Solution: 2015 Fall Final - 12
Author: Michiel SmidQuestion
A bowl contains 5 red balls and 7 blue balls. We choose a uniformly random subset of 3 balls. Define the event
- A = "exactly 2 of the chosen balls are red".
(a)
$\frac{5 \cdot {7 \choose 2}}{{12 \choose 3}}$
(b)
$\frac{{5 \choose 2} \cdot 7}{{12 \choose 3}}$
(c)
$\frac{{5 \choose 2}}{{12 \choose 3}}$
(d)
$\frac{{7 \choose 2}}{{12 \choose 3}}$
Solution
First, we choose 1 of the 7 blue balls: $ \binom{7}{1} = 7$
Then, we choose 2 of the 5 red balls: $ \binom{5}{2} $
The total number of ways to choose 3 balls from the 12 balls is $ \binom{12}{3} $
$ Pr(A) = \frac{7 \cdot \binom{5}{2}}{\binom{12}{3}} $