Solution: 2017 Winter Final - 24

Author: Michiel Smid

Question

Elisa Kazan has successfully completed her first term as President of the Carleton Computer Science Society. In order to celebrate this, Elisa decides to spend an evening in the Hyacintho Cactus Bar and Grill in downtown Ottawa. During this evening, Tan Tran is working as a server. Since Tan has been studying very hard for COMP 2804, he is a bit absent-minded: Every time a customer orders a drink, Tan serves the wrong drink with probability 1/12, independently of other orders.
Elisa orders 7 ciders, one cider at a time. Let $(D_1,D_2,\dots,D_7)$ be the sequence of drinks that Tan serves. Define the following random variable $X$:

X = the number of indices $i$ such that $D_i$ is a cider and $D_{i+1}$ is not a cider.

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

(a)

66/144

(b)

77/144

(c)

44/144

(d)

55/144

Solution

Thank God I have a second monitor to watch YouTube videos on

Alright, so let’s break it down. It’s asking for the probability that a $D_i$ is a cider and $D_{i+1}$ is not a cider

First has probability of being cider $ \frac{11}{12} $

Second has probability of not being cider $ \frac{1}{12} $

This occurs 6 times. Not 7 because $D_n$ has no $D_{n+1}$ to compare it to

$ \mathbb{E}(X) = \sum_{i=1}^{6} \frac{11}{12} \cdot \frac{1}{12} $

$ \mathbb{E}(X) = 6 \cdot \frac{11}{12} \cdot \frac{1}{12} $

$ \mathbb{E}(X) = 6 \cdot \frac{11}{144} $

$ \mathbb{E}(X) = \frac{6}{144} $