Question

You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win $8; if the faces are both heads, you win $4; if the coins do not match (one shows a head, the other a tail), you lose $2 (win (−$2)). Calculate the mean and variance of Y, your winnings on a single play of the game. Note that E(Y) > 0.

Answer 1

Answer:

Mean: $2

Variance: 18

Step-by-step explanation:

There are 4 possible outcomes:

TT (both tails)

TH (tail, head)

HT (head, tail)

HH (head, head)

Find the probabilities:

$P(TT)=P(TH)=P(HT)=P(HH)=\dfrac{1}{2}\cdot \dfrac{1}{2}=\dfrac{1}{4}$

Complete the table

$\begin{array}{ccccc}&TT&TH&HT&HH\\ \\P&\dfrac{1}{4}&\dfrac{1}{4}&\dfrac{1}{4}&\dfrac{1}{4}\\ \\Win&\ 8&-\ 2&-\ 2&\ 4\end{array}$

The mean is

$\dfrac{1}{4}\cdot 8+\dfrac{1}{4}\cdot (-2)+\dfrac{1}{4}\cdot (-2)+\dfrac{1}{4}\cdot 4=2$

The variance is

$\dfrac{(8-2)^2+(-2-2)^2+(-2-2)^2+(4-2)^2}{4}=\dfrac{36+16+16+4}{4}=18$