Question

karen has a bag with 8 balls numbered 1 through 8. she is playing a game of chance. this game is this: karen chooses one ball from the bag at random. she wins $1 if the number 1 is selected, $2 if the number 2 is selected, $3 if the number 3 is selected, $4 if the number 4 is selected, and $5 if the number 5 is selected. she loses $7 if 6, 7, or 8 is selected. (a) find the expected value of playing the game. dollars (b) what can karen expect in the long run, after playing the game many times? (she replaces the ball in the bag each time.) karen can expect to gain money. she can expect to win dollars per selection. karen can expect to lose money. she can expect to lose dollars per selection. karen can expect to break even (neither gain nor lose money).

Explanation:

Step1: Define probabilities and pay - offs

The probability of selecting each ball is $P(X = i)=\frac{1}{8}$ for $i = 1,2,\cdots,8$. The pay - offs are: $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, $x_4 = 4$, $x_5 = 5$, $x_6=-7$, $x_7=-7$, $x_8=-7$.

Step2: Use the expected - value formula

The expected value formula is $E(X)=\sum_{i = 1}^{n}x_iP(x_i)$.
\[

$$\begin{align*} E(X)&=(1\times\frac{1}{8})+(2\times\frac{1}{8})+(3\times\frac{1}{8})+(4\times\frac{1}{8})+(5\times\frac{1}{8})+(- 7\times\frac{3}{8})\\ &=\frac{1 + 2+3 + 4+5-21}{8}\\ &=\frac{15 - 21}{8}\\ &=-\frac{6}{8}\\ &=- 0.75 \end{align*}$$

\]

Answer:

(a) - 0.75
(b) Karen can expect to lose money. She can expect to lose 0.75 dollars per selection.