Solution: 2019 Winter Final - 19

Author: Michiel Smid

Question

A red box contains the numbers 0, 1, and 2; a blue box contains the numbers 0 and 1; and a green box contains the numbers 1 and 2. Consider the following two steps: Step 1: Choose a uniformly random number from the red box, and denote it by $x$.
Step 2:

If $x = 0$ or $x = 2$, then choose a uniformly random number from the green box, and denote it by $y$.
Otherwise (i.e., if $x = 1$), choose a uniformly random number from the blue box, and denote it by $y$.

Consider the random variables

X = the number $x$ you choose in Step 1,
Y = the number $y$ you choose in Step 2,
Z = $\max(X,Y).$

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

(a)

(b)

None of the above.

(c)

3/2

(d)

Solution

Honestly, let’s just write out all possible outcomes and their probabilities. It’s not that bad.

(0,1) with probability $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$
(0,2) with probability $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$
(1,0) with probability $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$
(1,1) with probability $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$
(2,1) with probability $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$
(2,2) with probability $\frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}$

$ \mathbb{E}(Z) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{3} + 1 \cdot \frac{1}{6} + 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} $

$ \mathbb{E}(Z) = \frac{9}{6} $

$ \mathbb{E}(Z) = \frac{3}{2} $