The spinner on a wheel of fortune can land with an equal chance on any one of ten regions.  Three regions are​ red, four are​ blue, two are​ yellow, and one is green.  A player wins 

$ 1$1 

if the spinner stops on red and 

$ 5$5 

if it stops on green.  The player loses 

$ 1$1 

if it stops on blue and 

$ 2$2 

if it stops on yellow.

Question content area bottom

Part 1

**a**. 

What is the expected​ value? 

​$0.000.00  

​(Round to the nearest​ cent.)

Part 2

**b.** What does this value mean if the game is played ten​ times?

A.

More than 10 games need to be played for the expected value to be a reasonable estimate of expected winnings.

B.

Over 10​ games, the player can expect to break even.

C.

Over 10​ games, the player can expect to lose money.

D.

Over 10​ games, the player can expect to win money.

To solve this problem, we will first calculate the expected value of a single game by considering the probability and outcome of each possible event. Then, we'll interpret this value in the context of playing ten games.

### Part 1: Expected Value Calculation

The expected value $ E(X) $ can be calculated using the formula:

\[
E(X) = \sum_{i} P(x_i) \times x_i
\]

where:
- $ P(x_i) $ is the probability of the outcome $ x_i $,
- $ x_i $ is the outcome (monetary value) for that event.

**Step 1: List the outcomes and their probabilities:**

1. **Red**:  
   - Probability: $ \frac{3}{10} $  
   - Outcome: +\$1

2. **Blue**:  
   - Probability: $ \frac{4}{10} $  
   - Outcome: -\$1

3. **Yellow**:  
   - Probability: $ \frac{2}{10} $  
   - Outcome: -\$2

4. **Green**:  
   - Probability: $ \frac{1}{10} $  
   - Outcome: +\$5

**Step 2: Calculate the expected value $ E(X) $:**

\[
E(X) = \left(\frac{3}{10} \times 1\right) + \left(\frac{4}{10} \times -1\right) + \left(\frac{2}{10} \times -2\right) + \left(\frac{1}{10} \times 5\right)
\]

\[
E(X) = \left(\frac{3}{10} \times 1\right) + \left(\frac{4}{10} \times -1\right) + \left(\frac{2}{10} \times -2\right) + \left(\frac{1}{10} \times 5\right)
\]

\[
E(X) = \frac{3}{10} - \frac{4}{10} - \frac{4}{10} + \frac{5}{10}
\]

\[
E(X) = \frac{3 - 4 - 4 + 5}{10} = \frac{0}{10} = 0
\]

### Final Answer for Part 1:
The expected value $ E(X) $ is **$0.00** (after rounding to the nearest cent).

### Part 2: Interpretation Over 10 Games

Since the expected value for one game is $0.00, over 10 games, the player can expect to **break even** on average. Therefore, the correct interpretation is:

**B. Over 10 games, the player can expect to break even.**

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### Questions:

1. What would the expected value be if the spinner had more red regions?
2. How would the expected value change if the payout for landing on green increased?
3. What is the expected value if all regions had equal payouts but different probabilities?
4. How does the variance in outcomes affect the reliability of the expected value?
5. How would the expected value change if there were more regions and outcomes?

**Tip:** When calculating expected values, consider both the probabilities and the outcomes—they determine whether the game is fair, favorable, or unfavorable.

The spinner on a wheel of fortune can land with an equal chance on any one of ten regions. Three regions are red, four are blue, two are yellow, and one is green. A player wins $1 if the spinner stops on red and $5 if it stops on green. The player loses $1 if it stops on blue and $2 if it stops on yellow. What is the expected value? What does this value mean if the game is played ten times?

Understand how to calculate the expected value of a wheel of fortune game with varying outcomes. This problem involves probability concepts and the expected value formula, suitable for students in Grades 9-12. Learn how the expected value reflects the average outcome over multiple plays.

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