Solution: 2018 Fall Final - 20

Author: Michiel Smid

Question

Let $n \geq 2$ be an integer. You are given $n$ beer bottles $B_1,B_2,\dots,B_n$ and two cider bottles $C_1$ and $C_2$. You choose a uniformly random 3-element subset of the set of these $n+2$ bottles. Define the random variable $X$ to be

X = the number of cider bottles in the chosen subset.

What is the expected value $\mathbb{E}(X)$ of the random variable $X$?

(a)

$\frac{2 {{n}\choose{2}} + 2n}{{{n+2}\choose{3}}}$

(b)

$\frac{2 {{n}\choose{2}} + n}{{{n+2}\choose{3}}}$

(c)

$\frac{2 {{n}\choose{2}} + n + 1}{{{n+2}\choose{3}}}$

(d)

$\frac{2 {{n}\choose{2}} + n - 1}{{{n+2}\choose{3}}}$

Solution

Let $X_i$ be 2 if there are 2 cider bottles in the subset, 1 if there is 1 cider bottle in the subset, and 0 if there are no cider bottles in the subset

$X_i = 0 $ if there are no cider bottles
First we choose 3 beer bottles: $ \binom{n}{3} $
Second we choose 0 cider bottles: $ \binom{2}{0} $
$ Pr(X_i = 0) = \frac{ \binom{n}{3} \cdot \binom{2}{0} }{ \binom{n+2}{3} } $
$X_i = 1 $ if there is 1 cider bottle
First we choose 1 cider bottle: $ \binom{2}{1} $
Second we choose 2 beer bottles: $ \binom{n}{2} $
$ Pr(X_i = 1) = \frac{ \binom{2}{1} \cdot \binom{n}{2} }{ \binom{n+2}{3} } $
$X_i = 2 $ if there are 2 cider bottles
First we choose 2 cider bottles: $ \binom{2}{2} $
Second we choose 1 beer bottle: $ \binom{n}{1} $
$ Pr(X_i = 2) = \frac{ \binom{2}{2} \cdot \binom{n}{1} }{ \binom{n+2}{3} } $

$ \mathbb{E}(X) = 0 \cdot Pr(X_i = 0) + 1 \cdot Pr(X_i = 1) + 2 \cdot Pr(X_i = 2) $

$ \mathbb{E}(X) = 0 \cdot \frac{ \binom{n}{3} \cdot \binom{2}{0} }{ \binom{n+2}{3} } + 1 \cdot \frac{ \binom{2}{1} \cdot \binom{n}{2} }{ \binom{n+2}{3} } + 2 \cdot \frac{ \binom{2}{2} \cdot \binom{n}{1} }{ \binom{n+2}{3} } $

$ \mathbb{E}(X) = \frac{ \binom{2}{1} \cdot \binom{n}{2} }{ \binom{n+2}{3} } + 2 \cdot \frac{ \binom{2}{2} \cdot \binom{n}{1} }{ \binom{n+2}{3} } $

$ \mathbb{E}(X) = \frac{2 \cdot \binom{n}{2} + 2 \cdot \binom{n}{1} }{ \binom{n+2}{3} } $

$ \mathbb{E}(X) = \frac{2 \cdot \binom{n}{2} + 2 \cdot n }{ \binom{n+2}{3} } $