Answer :
Certainly! Let's tackle the problem step by step, providing clear and detailed calculations.
### Part a: Fill out the missing values in the table
There are various colored sectors on the wheel, and each color has a specific point value attached to it:
- Blue: 1 sector, 3 points
- Yellow: 2 sectors, 1 point each
- Purple: 2 sectors, 0 points each
- Red: 2 sectors, -1 point each
Here's the summary table:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Color} & \text{Number of Sectors} & \text{Points} \\ \hline \text{Blue} & 1 & 3 \\ \hline \text{Yellow} & 2 & 1 \\ \hline \text{Purple} & 2 & 0 \\ \hline \text{Red} & 2 & -1 \\ \hline \end{array} \][/tex]
### Part b: Calculate the Expected Value of a Single Spin
The expected value [tex]\( E(X) \)[/tex] of a random variable [tex]\( X \)[/tex] is computed as follows:
[tex]\[ E(X) = \sum (P(X = x) \cdot x) \][/tex]
Where [tex]\( P(X = x) \)[/tex] is the probability of getting [tex]\( x \)[/tex] points.
The probabilities of landing on each sector color are:
[tex]\[ P(\text{Blue}) = \frac{1}{7}, \quad P(\text{Yellow}) = \frac{2}{7}, \quad P(\text{Purple}) = \frac{2}{7}, \quad P(\text{Red}) = \frac{2}{7} \][/tex]
The expected value calculation involves summing up the product of each outcome's probability and its associated points:
[tex]\[ \begin{aligned} E(X) &= P(\text{Blue}) \times \text{Points}(\text{Blue}) + P(\text{Yellow}) \times \text{Points}(\text{Yellow}) + P(\text{Purple}) \times \text{Points}(\text{Purple}) + P(\text{Red}) \times \text{Points}(\text{Red}) \\ &= \left( \frac{1}{7} \times 3 \right) + \left( \frac{2}{7} \times 1 \right) + \left( \frac{2}{7} \times 0 \right) + \left( \frac{2}{7} \times -1 \right) \\ &= \frac{3}{7} + \frac{2}{7} + 0 - \frac{2}{7} \\ &= \frac{3}{7} + 0 + 0 - \frac{2}{7} \\ &= \frac{3 - 2}{7} \\ &= \frac{1}{7} \end{aligned} \][/tex]
So, the expected value for a single spin is approximately:
[tex]\[ E(X) \approx 0.4286 \][/tex]
### Part c: Make the Game Fair
A fair game has an expected value of 0. To make the game fair, we need to adjust the point values such that the expected value equals 0.
Given the current number of 1 blue sector, 2 yellow sectors, 2 purple sectors, and 2 red sectors, we need to determine new point values [tex]\( p_{\text{red}} \)[/tex] for the game to have a neutral expected value (0).
Retaining the current point values for blue, yellow, and purple:
[tex]\[ \text{Points}(\text{Blue}) = 3, \quad \text{Points}(\text{Yellow}) = 1, \quad \text{Points}(\text{Purple}) = 0 \][/tex]
Let [tex]\( p_{\text{red}} \)[/tex] be the new point value for a red sector. Setting up the equation for expected value 0:
[tex]\[ \left( \frac{1}{7} \times 3 \right) + \left( \frac{2}{7} \times 1 \right) + \left( \frac{2}{7} \times 0 \right) + \left( \frac{2}{7} \times p_{\text{red}} \right) = 0 \][/tex]
Solving for [tex]\( p_{\text{red}} \)[/tex]:
[tex]\[ \frac{3}{7} + \frac{2}{7} \times 1 + \frac{2}{7} \times 0 + \frac{2}{7} \times p_{\text{red}} = 0 \][/tex]
[tex]\[ \frac{3}{7} + \frac{2}{7} + \frac{2}{7} \times p_{\text{red}} = 0 \][/tex]
[tex]\[ \frac{3 + 2 + 2 \cdot p_{\text{red}}}{7} = 0 \][/tex]
[tex]\[ \frac{5 + 2 \cdot p_{\text{red}}}{7} = 0 \][/tex]
[tex]\[ 5 + 2 \cdot p_{\text{red}} = 0 \][/tex]
[tex]\[ 2 \cdot p_{\text{red}} = -5 \][/tex]
[tex]\[ p_{\text{red}} = -\frac{5}{2} \][/tex]
Therefore, the new point value for the red sector should be:
[tex]\[ p_{\text{red}} = -2.5 \][/tex]
Let's verify this new point value results in a fair game:
[tex]\[ \begin{aligned} &\left( \frac{1}{7} \times 3 \right) + \left( \frac{2}{7} \times 1 \right) + \left( \frac{2}{7} \times 0 \right) + \left( \frac{2}{7} \times -2.5 \right) \\ &= \frac{3}{7} + \frac{2}{7} \times 1 + 0 + \frac{2}{7} \times -2.5 \\ &= \frac{3}{7} + \frac{2}{7} \times 1 + \frac{2}{7} \times -2.5 \\ &= \frac{3}{7} + \frac{2}{7} - \frac{5}{7} \\ &= \frac{3 + 2 - 5}{7} \\ &= \frac{0}{7} \\ &= 0 \end{aligned} \][/tex]
Thus, the game is fair when the red sector's points are adjusted to -2.5.
### Part a: Fill out the missing values in the table
There are various colored sectors on the wheel, and each color has a specific point value attached to it:
- Blue: 1 sector, 3 points
- Yellow: 2 sectors, 1 point each
- Purple: 2 sectors, 0 points each
- Red: 2 sectors, -1 point each
Here's the summary table:
[tex]\[ \begin{array}{|c|c|c|} \hline \text{Color} & \text{Number of Sectors} & \text{Points} \\ \hline \text{Blue} & 1 & 3 \\ \hline \text{Yellow} & 2 & 1 \\ \hline \text{Purple} & 2 & 0 \\ \hline \text{Red} & 2 & -1 \\ \hline \end{array} \][/tex]
### Part b: Calculate the Expected Value of a Single Spin
The expected value [tex]\( E(X) \)[/tex] of a random variable [tex]\( X \)[/tex] is computed as follows:
[tex]\[ E(X) = \sum (P(X = x) \cdot x) \][/tex]
Where [tex]\( P(X = x) \)[/tex] is the probability of getting [tex]\( x \)[/tex] points.
The probabilities of landing on each sector color are:
[tex]\[ P(\text{Blue}) = \frac{1}{7}, \quad P(\text{Yellow}) = \frac{2}{7}, \quad P(\text{Purple}) = \frac{2}{7}, \quad P(\text{Red}) = \frac{2}{7} \][/tex]
The expected value calculation involves summing up the product of each outcome's probability and its associated points:
[tex]\[ \begin{aligned} E(X) &= P(\text{Blue}) \times \text{Points}(\text{Blue}) + P(\text{Yellow}) \times \text{Points}(\text{Yellow}) + P(\text{Purple}) \times \text{Points}(\text{Purple}) + P(\text{Red}) \times \text{Points}(\text{Red}) \\ &= \left( \frac{1}{7} \times 3 \right) + \left( \frac{2}{7} \times 1 \right) + \left( \frac{2}{7} \times 0 \right) + \left( \frac{2}{7} \times -1 \right) \\ &= \frac{3}{7} + \frac{2}{7} + 0 - \frac{2}{7} \\ &= \frac{3}{7} + 0 + 0 - \frac{2}{7} \\ &= \frac{3 - 2}{7} \\ &= \frac{1}{7} \end{aligned} \][/tex]
So, the expected value for a single spin is approximately:
[tex]\[ E(X) \approx 0.4286 \][/tex]
### Part c: Make the Game Fair
A fair game has an expected value of 0. To make the game fair, we need to adjust the point values such that the expected value equals 0.
Given the current number of 1 blue sector, 2 yellow sectors, 2 purple sectors, and 2 red sectors, we need to determine new point values [tex]\( p_{\text{red}} \)[/tex] for the game to have a neutral expected value (0).
Retaining the current point values for blue, yellow, and purple:
[tex]\[ \text{Points}(\text{Blue}) = 3, \quad \text{Points}(\text{Yellow}) = 1, \quad \text{Points}(\text{Purple}) = 0 \][/tex]
Let [tex]\( p_{\text{red}} \)[/tex] be the new point value for a red sector. Setting up the equation for expected value 0:
[tex]\[ \left( \frac{1}{7} \times 3 \right) + \left( \frac{2}{7} \times 1 \right) + \left( \frac{2}{7} \times 0 \right) + \left( \frac{2}{7} \times p_{\text{red}} \right) = 0 \][/tex]
Solving for [tex]\( p_{\text{red}} \)[/tex]:
[tex]\[ \frac{3}{7} + \frac{2}{7} \times 1 + \frac{2}{7} \times 0 + \frac{2}{7} \times p_{\text{red}} = 0 \][/tex]
[tex]\[ \frac{3}{7} + \frac{2}{7} + \frac{2}{7} \times p_{\text{red}} = 0 \][/tex]
[tex]\[ \frac{3 + 2 + 2 \cdot p_{\text{red}}}{7} = 0 \][/tex]
[tex]\[ \frac{5 + 2 \cdot p_{\text{red}}}{7} = 0 \][/tex]
[tex]\[ 5 + 2 \cdot p_{\text{red}} = 0 \][/tex]
[tex]\[ 2 \cdot p_{\text{red}} = -5 \][/tex]
[tex]\[ p_{\text{red}} = -\frac{5}{2} \][/tex]
Therefore, the new point value for the red sector should be:
[tex]\[ p_{\text{red}} = -2.5 \][/tex]
Let's verify this new point value results in a fair game:
[tex]\[ \begin{aligned} &\left( \frac{1}{7} \times 3 \right) + \left( \frac{2}{7} \times 1 \right) + \left( \frac{2}{7} \times 0 \right) + \left( \frac{2}{7} \times -2.5 \right) \\ &= \frac{3}{7} + \frac{2}{7} \times 1 + 0 + \frac{2}{7} \times -2.5 \\ &= \frac{3}{7} + \frac{2}{7} \times 1 + \frac{2}{7} \times -2.5 \\ &= \frac{3}{7} + \frac{2}{7} - \frac{5}{7} \\ &= \frac{3 + 2 - 5}{7} \\ &= \frac{0}{7} \\ &= 0 \end{aligned} \][/tex]
Thus, the game is fair when the red sector's points are adjusted to -2.5.
