Solution: 2018 Fall Final - 13

Author: Michiel Smid

Question

You are given a fair die that has six faces. One face has the letter $a$ on it, two faces have the letter $b$ on them, and three faces have the letter $c$ on them. Assume you roll this die twice, independently of each other. Define the events

A = "both rolls result in the same letter",
B = "at least one of the rolls results in the letter $a$".

What is $\Pr(A|B)$?

(a)

2/11

(b)

1/7

(c)

1/11

(d)

2/7

Solution

In this case, our dice has special properties compared to a normal 6-sided dice.
We can define the dice as having the following results: {a, b, b, c, c, c}
Now, all we have to do is determine $Pr(A \cap B)$ and $Pr(B)$ to determine $Pr(A|B)$ using the independence formula.

$Pr(B)$ represents the probability of at least one of the rolls resulting in the letter $a$. To make this easier to visualize, we can draw out a table of all possible outcomes between the two dice rolls.
$|S|$ = the sample space of all possible rolls = $6 \times 6 = 36$

All Possible Rolls for Two Custom 6-Sided Dice
	Die 2: a	Die 2: b	Die 2: b	Die 2: c	Die 2: c	Die 2: c
Die 1: a	(a,a)	(a,b)	(a,b)	(a,c)	(a,c)	(a,c)
Die 1: b	(b,a)	(b,b)	(b,b)	(b,c)	(b,c)	(b,c)
Die 1: c	(c,a)	(c,b)	(c,b)	(c,c)	(c,c)	(c,c)

As we can see from the table, there are 11 possible outcomes that have at least 1 $a$ in them. Therefore, $Pr(B) = \frac{|B|}{|S|} = \frac{11}{36}$.

For $Pr(A \cap B)$, we only care about the cases when both rolls result in the same letter AND at least one of the rolls result in the letter $a$.
From the table, we can see that there is only 1 possible outcome that satisfies both conditions, which is the case where both dice roll an $a$. Therefore, $Pr(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{36}$.

Given this, we can now calculate $Pr(A|B)$ using the independence formula:

$ Pr(A|B) = \frac{ Pr(A \cap B) }{ Pr(B) } = \frac{ \frac{1}{36} }{ \frac{11}{36} } = \frac{1}{11} $