Solution: 2018 Winter Final - 24

Author: Michiel Smid

Question

Elisa Kazan has successfully completed her second year as President of the Carleton Computer Science Society. In order to celebrate this, Elisa throws a party. She invites 15 students; thus, the total number of students at the party is equal to 16. Elisa has brought an unlimited amount of drinks: 5 types $C_1,C_2,C_3,C_4,C_5$ of cider and 3 types $B_1,B_2,B_3$ of beer. Each of the 16 students gets 3 drinks; each of these drinks is uniformly, and independently, chosen from the 8 types of drinks.
Define the following random variable $X$:

X = the number of students who get exactly 2 ciders.

What is the expected value $\mathbb{E}(X)$ of the random variable $X$?
Hint: Use indicator random variables.

(a)

$2^{4} \cdot 3^{2} \left. \cdot 5^{2} \middle/ 8^{3} \right.$

(b)

$2^{4} \cdot 3^{2} \cdot \left. 2^{5} \middle/ 8^{3} \right.$

(c)

$3^{2} \cdot \left. 5^{2} \middle/ 8^{3} \right.$

(d)

$2^{4} \cdot 3^{2} \cdot \left. 5^{2} \middle/ 3^{8} \right.$

Solution

Let S be the set of all possibilities = $ |S| = 8^{3} $

Let A be the set of all possibilities where a student gets exactly 2 ciders

To get exactly 2 ciders, a student has to have 2 ciders and 1 beer. There are 3 ways to get 2 ciders and a beer: $CBC, BCC, CCB$.

Number of ways for combination $CBC$: $\binom{5}{1} \cdot \binom{3}{1} \cdot \binom{5}{1} = 5^2 \cdot 3$
Number of ways for combination $BCC$: $\binom{3}{1} \cdot \binom{5}{1} \cdot \binom{5}{1} = 5^2 \cdot 3$
Number of ways for combination $CCB$: $\binom{5}{1} \cdot \binom{5}{1} \cdot \binom{3}{1} = 5^2 \cdot 3$

Now, we can sum

$ Pr(A) = Pr( CBC ) + Pr( BCC ) + Pr( CCB ) $

$ Pr(A) = (5^2 \cdot 3) + (5^2 \cdot 3) + (5^2 \cdot 3) $

$ Pr(A) = 3 \cdot \frac{5^2 \cdot 3}{8^3} $

$ Pr(A) = \frac{3^2 \cdot 5^2}{8^3}$

Let $X_i$ be an indicator random variable where:

$ 1 \text{ if a student gets exactly 2 ciders} $

$ 0 \text{ otherwise} $

$ Pr(X_i = 1) = Pr(A) = \frac{3^2 \cdot 5^2}{8^3}$

$ \mathbb{E}(X) = \sum_{k=1}^{16} 1 \cdot Pr(X_i = 1) $

$ \mathbb{E}(X) = \sum_{k=1}^{16} \frac{3^2 \cdot 5^2}{8^3} $

$ \mathbb{E}(X) = 16 \cdot \frac{3^2 \cdot 5^2}{8^3} $

$ \mathbb{E}(X) = 2^4 \cdot \frac{3^2 \cdot 5^2}{8^3} $

$ \mathbb{E}(X) = \frac{2^4 \cdot 3^2 \cdot 5^2}{8^3} $

$ \mathbb{E}(X) = 2^4 \cdot 3^2 \cdot \frac{5^2}{8^3} $