Solution: 2018 Fall Final - 18

Author: Michiel Smid

Question

A red box contains the numbers 0, 1, and 2, and a blue box also contains the numbers 0, 1, and 2. You choose a uniformly random element from the red box and a uniformly random element from the blue box; these two choices are independent of each other. Define the random variables

X = the number you choose from the red box,
Y = the number you choose from the blue box,
Z = $\max(X, Y)$.

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

(a)

9/13

(b)

8/9

(c)

13/9

(d)

9/8

Solution

Brute force it with all posibilities

$ { (00), (01), (02), (10), (11), (12), (20), (21), (22) } $

So straight up, there are 9 possibilities and the probability of each ordered pair has a chance of occuring with $ \frac{1}{9} $

In the order that the orered pairs appear above, I’m gonna write down the max of each pair along with the probability of that pair occuring

$ \frac{1}{9} \cdot 0 + \frac{1}{9} \cdot 1 + \frac{1}{9} \cdot 2 + \frac{1}{9} \cdot 1 + \frac{1}{9} \cdot 1 + \frac{1}{9} \cdot 2 + \frac{1}{9} \cdot 2 + \frac{1}{9} \cdot 2 + \frac{1}{9} \cdot 2 $

$ = \frac{1}{9} ( 0 + 1 + 2 + 1 + 1 + 2 + 2 + 2 + 2 ) $

$ = \frac{1}{9} ( 13 ) $

$ = \frac{13}{9} $