Solution: 2015 Fall Final - 17

Author: Michiel Smid

Question

You are given that:

The course COMP 9999 runs over a period of one year, starting on January 1 and ending on December 31. There is one lecture every day; thus, the total number of lectures is 365.
At the beginning of each of the 365 lectures, Nick flips a fair and independent coin twice. If the coin comes up heads twice, then Nick eats 3 bananas during the lecture; otherwise, Nick eats 5 bananas during the lecture.

Let $X$ be the total number of bananas that Nick eats during the 365 lectures of the course COMP 9999. What is the expected value $\mathbb{E}(X)$ of $X$?
(n.b., you may find it useful to apply Linearity of Expectation)

(a)

$\frac{7 \cdot 365}{2}$

(b)

$\frac{9 \cdot 365}{2}$

(c)

$\frac{5 \cdot 365}{2}$

(d)

$4 \cdot 365$

Solution

Let $Y_i$ be 1 if Nick gets 2 heads and 0 otherwise.

The chances of getting 2 heads is $ \frac{1}{4} $

$E(Y_i=1) = \frac{1}{4} $

Let $Z_i$ be 1 if Nick does not get 2 heads and 0 otherwise.

The chances of getting 1 or 0 heads is $ \frac{3}{4} $

$E(Z_i=1) = \frac{3}{4} $

$ X = \sum_{i=1}^{365} 3 \cdot E(Y_i) + \sum_{i=1}^{365} 5 \cdot E(Z_i) $

$ X = 3 \cdot 365 \cdot \frac{1}{4} + 5 \cdot 365 \cdot \frac{3}{4} $

$ X = \frac{3 \cdot 365}{4} + \frac{5 \cdot 365 \cdot 3}{4} $

$ X = \frac{3 \cdot 365 + 5 \cdot 365 \cdot 3}{4} $

$ X = \frac{3 \cdot 365 + 15 \cdot 365}{4} $

$ X = \frac{ (3+15) \cdot 365}{4} $

$ X = \frac{18 \cdot 365}{4} $

$ X = \frac{9 \cdot 365}{2} $