Solution: 2014 Winter Final - 16

Author: Michiel Smid

Question

Let $C_1$ be a fair coin that has $H$ on one side and $T$ on the other side. Let $C_2$ be a coin that has $H$ on both sides. We choose one of $C_1$ and $C_2$ uniformly at random and flip it. Define the events

A = "we choose $C_2$",
B = "the flip resulted in $H$".

What is the conditional probability $\Pr(A|B)$?

(a)

none of the above

(b)

1/3

(c)

2/3

(d)

1/2

Solution

Let S be the set of all outcomes in terms of coin chosen and face flipped.
The set of S is $ { (C_1, H), (C_1, T), (C_2, H), (C_2, H) } $
The size of S is the number of outcomes: $ |S| = 4 $
Let's find the probability of A
The set of A is $ { (C_2, H), (C_2, H) } $
The size of A is the number of outcomes in $A$: $ |A| = 2 $
$ Pr(A) = \frac{|A|}{|S|} = \frac{2}{4} = \frac{1}{2} $
Let's find the probability of B
The set of B is $ { (C_1, H), (C_2, H), (C_2, H) } $
The size of B is the number of outcomes in $B$: $ |B| = 3 $
$ Pr(B) = \frac{|B|}{|S|} = \frac{3}{4} $
Let's find the probability of $A \cap B$
The set of $A \cap B$ is $ { (C_2, H), (C_2, H) } $
The size of $A \cap B$ is the number of outcomes in $A \cap B$: $ |A \cap B| = 2 $
$ Pr(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{2}{4} = \frac{1}{2} $

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

$ Pr(A|B) = \frac{ \frac{1}{2}}{ \frac{3}{4}} $

$ Pr(A|B) = \frac{1}{2} \times \frac{4}{3} $

$ Pr(A|B) = \frac{2}{3} $