Solution: 2015 Fall Midterm - 17
Author: Michiel SmidQuestion
Let $S = \{1,2,3,4,5,6,7\}$. You choose a uniformly random 3-element subset $X$ of $S$. Thus,
each 3-element subset of $S$ has a probability of $\left. 1 \middle/ {7 \choose 3} \right.$ of being $X$.
Define the event
- A = "4 is an element of $X$".
(a)
7/3
(b)
15/7
(c)
7/15
(d)
3/7
Solution
Assuming we obtained a 4 in our 3-element subset, we have 2 more spots to fill.
Thus, there are $\binom{6}{2}$ ways to fill the remaining spots.
$|A| = \binom{6}{2}$
$ Pr(A) = \frac{\binom{6}{2}}{\binom{7}{3}} $
$ Pr(A) = \frac{ \frac{6!}{2!4!}}{ \frac{7!}{3!4!}}$
$ Pr(A) = \frac{6!}{2!4!} \cdot \frac{3!4!}{7!}$
$ Pr(A) = \frac{1}{2!4!} \cdot \frac{3!4!}{7}$
$ Pr(A) = \frac{1}{2!} \cdot \frac{3!}{7}$
$ Pr(A) = \frac{1}{1} \cdot \frac{3}{7}$
$ Pr(A) = \frac{3}{7}$