To solve the problems presented in the Student Guide, we will break down each scenario step-by-step, calculating probabilities and expected values where required.
Miguel's Game Scenario
Step 1: Complete the Probability Table
- Chips: 2 chips with number 1, 1 chip with number 3, and 1 chip with number 5.
- Outcomes:
- Winning $2: Miguel picks 2 chips with the number 1.
- Losing $1: Miguel picks one of each of the different number chips (1 and 3, 1 and 5, 3 and 5).
Let’s find the outcomes:
- Picking (1, 1) → Wins $2
- Picking (1, 3) → Loses $1
- Picking (1, 5) → Loses $1
- Picking (3, 5) → Loses $1
Probability Calculation:
- Total ways to pick 2 chips from 4 = C(4, 2) = 6
- Ways to win (1, 1) = 1
- Ways to lose (1, 3) = 2
- Ways to lose (1, 5) = 2
- Ways to lose (3, 5) = 1
Table Format:
| Xi | P(xi) | |------|-------| | 2 | 1/6 | | -1 | 5/6 |
Step 2: Calculate Expected Value (E(X))
The expected value is calculated using the formula:
\[ E(X) = \sum (X_i \cdot P(X_i)) \]
Where:
- \( E(X) = 2 \cdot \frac{1}{6} + (-1) \cdot \frac{5}{6} \)
Calculation:
\[ E(X) = \frac{2}{6} - \frac{5}{6} = \frac{2 - 5}{6} = \frac{-3}{6} = -0.5 \]
Expected Value: -$0.50
Conclusion: Miguel should expect to lose $0.50 each time he plays.
Step 3: Making the Game Fair
To make the game fair (expected value of zero), if Miguel needs to adjust the winnings for choosing two chips with the number 1, let’s denote the new win as \( Y \).
To find \( Y \), we set up the equation:
\[ Y \cdot \frac{1}{6} + (-1) \cdot \frac{5}{6} = 0 \]
Solving for \( Y \):
\[ Y \cdot \frac{1}{6} = \frac{5}{6} \]
\[ Y = 5 \]
To make the game fair, the payout for choosing the two chips with the number 1 should be $5.
Spin a Wheel Game
Step 1: Complete the Probability Table
- Sectors: 2 Red, 2 Purple, 2 Yellow, 1 Blue
- Points for each color:
- Blue: 3 points
- Yellow: 1 point
- Purple: 0 points
- Red: -1 points
Probability Table:
| Xi | Points | |----|--------| | 3 | 1/7 | | 1 | 2/7 | | 0 | 2/7 | | -1 | 2/7 |
Step 2: Calculate Expected Value
\[ E(X) = 3 \cdot \frac{1}{7} + 1 \cdot \frac{2}{7} + 0 \cdot \frac{2}{7} + (-1) \cdot \frac{2}{7} \]
Calculation:
\[ E(X) = \frac{3}{7} + \frac{2}{7} + 0 - \frac{2}{7} = \frac{3}{7} \]
Expected Value: \( \frac{3}{7} \approx 0.43 \)
Step 3: Make the Game Fair
To make the game fair, the expected value should equal 0. Let's consider adjusting the values assigned to the outcomes:
Suppose we change the payout for the red sectors so that the expected value equals 0.
Let’s denote the modified payout for the red sector as \( R \):
Setting up the equation for fairness: \[ 3 \cdot \frac{1}{7} + 1 \cdot \frac{2}{7} + 0 \cdot \frac{2}{7} + R \cdot \frac{2}{7} = 0 \] This simplifies to: \[ \frac{3}{7} + \frac{2}{7} + R \cdot \frac{2}{7} = 0 \] Calculating for \( R \): \[ R \cdot \frac{2}{7} = -\frac{5}{7} \implies R = -\frac{5}{2} \Rightarrow -2.5 \]
This would imply that if landing on the red sector costs the player $2.50, the game would become fair.
Basketball Player Decision Scenario
Step 1: Calculate Expected Values
Three-point shot:
| Xi | P(xi) | |----|-------| | 3 | 0.30 | | 0 | 0.70 |
\[ E(X_{3pt}) = 3 \cdot 0.30 + 0 \cdot 0.70 = 0.9 \]
Two-point shot:
| Xi | P(xi) | |----|-------| | 2 | 0.48 | | 0 | 0.52 |
\[ E(X_{2pt}) = 2 \cdot 0.48 + 0 \cdot 0.52 = 0.96 \]
Conclusion:
The expected value of taking the three-point shot is $0.90, while passing the ball for the two-point shot has an expected value of $0.96. Therefore, the player should pass the ball as it has a higher expected value.
Claire's Investment
Step 1: Calculate Expected Value
Let’s denote outcomes:
- Lose $10,000: P = 0.2
- Break even $0: P = 0.4
- Profit $5,000: P = 0.3
- Profit $8,000: P = 0.1
The expected value is:
\[ E(X) = (-10,000) \cdot 0.2 + 0 \cdot 0.4 + 5,000 \cdot 0.3 + 8,000 \cdot 0.1 \]
Calculating:
\[ E(X) = -2000 + 0 + 1500 + 800 = -700 \]
Conclusion: Since the expected value is -$700, Claire should not invest in the company as she is expected to lose money.
Investment Recovery Time
To calculate the number of years to recover the initial investment of $1,200:
- With an expected loss of $700 per year (negative expected value), Claire would never earn back the money.
Thus, if the expected value remains constant, Claire will never earn back her initial investment.
Tanya's Game Scenario
Step 1: Calculate Expected Values for Each Game
Game 1:
- Lose $2: P = 0.55
- Win $1: P = 0.20
- Win $4: P = 0.25
\[ E(X) = (-2) \cdot 0.55 + 1 \cdot 0.20 + 4 \cdot 0.25 \] \[ E(X) = -1.1 + 0.2 + 1 = 0.1 \]
Game 2:
- Lose $2: P = 0.15
- Win $1: P = 0.35
- Win $4: P = 0.50
\[ E(X) = (-2) \cdot 0.15 + 1 \cdot 0.35 + 4 \cdot 0.50 \] \[ E(X) = -0.30 + 0.35 + 2 = 2.05 \]
Game 3:
- Lose $2: P = 0.20
- Win $1: P = 0.60
- Win $4: P = 0.20
\[ E(X) = (-2) \cdot 0.20 + 1 \cdot 0.60 + 4 \cdot 0.20 \] \[ E(X) = -0.40 + 0.60 + 0.80 = 1.0 \]
Conclusion:
- Game 1 Expected Value: $0.10
- Game 2 Expected Value: $2.05
- Game 3 Expected Value: $1.00
Tanya should choose Game 2 to maximize her expected value with a potential gain of $2.05.
This concludes the solutions needed for the tasks outlined in the Student Guide, demonstrating the calculations, reasoning, and expected outcomes for each scenario described.
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