Question

diane has a bag with 8 balls numbered 1 through 8. she is playing a game of chance.
this game is this: diane chooses one ball from the bag at random. she wins $1 if the number 1 is selected, $2 if the number 2 is selected, $5 if the number 3 is selected, $6 if the number 4 is selected, $8 if the number 5 is selected, and $10 if the number 6 is selected. she loses $17 if 7 or 8 is selected.

(a) find the expected value of playing the game.
$square$ dollars

(b) what can diane expect in the long run, after playing the game many times?
(she replaces the ball in the bag each time.)

$\bigcirc$ diane can expect to gain money.
she can expect to win $square$ dollars per selection.

$\bigcirc$ diane can expect to lose money.
she can expect to lose $square$ dollars per selection.

$\bigcirc$ diane can expect to break even (neither gain nor lose money).

Explanation:

Part (a)

Step 1: Determine Probabilities

There are 8 balls, so the probability of selecting any single ball is $\frac{1}{8}$. For numbers 1 - 6, each has a probability of $\frac{1}{8}$, and for numbers 7 and 8, together they have a probability of $\frac{2}{8}=\frac{1}{4}$ (but each of 7 and 8 has $\frac{1}{8}$).

Step 2: List Outcomes and Values

• Outcome 1: Value = $1$, Probability = $\frac{1}{8}$
• Outcome 2: Value = $2$, Probability = $\frac{1}{8}$
• Outcome 3: Value = $5$, Probability = $\frac{1}{8}$
• Outcome 4: Value = $6$, Probability = $\frac{1}{8}$
• Outcome 5: Value = $8$, Probability = $\frac{1}{8}$
• Outcome 6: Value = $10$, Probability = $\frac{1}{8}$
• Outcome 7: Value = $-17$ (loses), Probability = $\frac{1}{8}$
• Outcome 8: Value = $-17$ (loses), Probability = $\frac{1}{8}$

Step 3: Calculate Expected Value

The formula for expected value $E(X)$ is $E(X)=\sum [x_i \cdot P(x_i)]$, where $x_i$ is the value of the outcome and $P(x_i)$ is its probability.

\[

$$\begin{align*} E(X)&=(1\times\frac{1}{8})+(2\times\frac{1}{8})+(5\times\frac{1}{8})+(6\times\frac{1}{8})+(8\times\frac{1}{8})+(10\times\frac{1}{8})+(-17\times\frac{1}{8})+(-17\times\frac{1}{8})\\ &=\frac{1 + 2 + 5 + 6 + 8 + 10 - 17 - 17}{8}\\ &=\frac{(1+2+5+6+8+10)-(17+17)}{8}\\ &=\frac{32 - 34}{8}\\ &=\frac{-2}{8}\\ &=-\frac{1}{4}\\ &=-0.25 \end{align*}$$

\]

The expected value from part (a) is -0.25, which is negative. A negative expected value means that in the long run, Diane can expect to lose money. The magnitude of the expected value is the amount she can expect to lose per selection.

Answer:

-0.25

Part (b)