Answer :
Certainly! Let's go through each part of the problem step by step.
### a. Filling out the Table
Miguel is playing a game where:
1. He wins \[tex]$2 if he draws two chips with the same number. 2. He loses \$[/tex]1 if he draws two chips with different numbers.
Since there are four chips in the box with numbers {1, 1, 3, 5}, the possible outcomes for drawing two chips are:
- (1, 1)
- (1, 3)
- (1, 5)
- (1, 3)
- (1, 5)
- (3, 5)
Out of these 6 possible outcomes:
- Only 1 of them is a win (both chips being the same) leading to a gain of \[tex]$2: (1, 1). - The remaining 5 outcomes are losses (different numbers) leading to a loss of \$[/tex]1.
Thus, the probability table is:
[tex]\[ \begin{array}{|c|c|c|} \hline X_1 & 2 & -1 \\ \hline P(X_1) & \frac{1}{6} & \frac{5}{6} \\ \hline \end{array} \][/tex]
### b. Miguel's Expected Value
The expected value [tex]\( E(X) \)[/tex] is calculated by multiplying each outcome [tex]\( X_i \)[/tex] by its probability [tex]\( P(X_i) \)[/tex] and summing these products:
[tex]\[ E(X) = X_1 \cdot P(X_1) + X_2 \cdot P(X_2) \][/tex]
Substituting the given values:
[tex]\[ E(X) = 2 \cdot \frac{1}{6} + (-1) \cdot \frac{5}{6} \][/tex]
This calculation simplifies to:
[tex]\[ E(X) = \frac{2}{6} - \frac{5}{6} = -\frac{3}{6} = -0.5 \][/tex]
Thus, the expected value [tex]\( E(X) \)[/tex] is [tex]\(-0.5\)[/tex].
### c. Expected Money Won or Lost Each Time
Based on the expected value calculated in the previous step, Miguel should expect to lose on average [tex]\(-\$0.5\)[/tex] (50 cents) each time he plays the game. This outcome means that over a large number of games, Miguel will lose 50 cents per game on average.
### d. Making the Game Fair
To make the game fair, the expected value [tex]\( E(X) \)[/tex] should be zero. Let [tex]\( Y \)[/tex] be the new value assigned for the case when Miguel wins by choosing two chips with the number 1.
The new equation for a fair game will be:
[tex]\[ Y \cdot P(X_1) + (-1) \cdot P(X_2) = 0 \][/tex]
Substituting the probabilities:
[tex]\[ Y \cdot \frac{1}{6} + (-1) \cdot \frac{5}{6} = 0 \][/tex]
Solving for [tex]\( Y \)[/tex]:
[tex]\[ \frac{Y}{6} - \frac{5}{6} = 0 \][/tex]
[tex]\[ \frac{Y}{6} = \frac{5}{6} \][/tex]
Multiplying both sides by 6:
[tex]\[ Y = 5 \][/tex]
Therefore, to make the game fair, the value that should be assigned to picking two chips with the number 1 should be [tex]\( \$5 \)[/tex].
In summary:
- The expected value is [tex]\(-0.5\)[/tex].
- Miguel should expect to lose 50 cents each game.
- To make the game fair, winning with two number 1 chips should be worth \$5.
### a. Filling out the Table
Miguel is playing a game where:
1. He wins \[tex]$2 if he draws two chips with the same number. 2. He loses \$[/tex]1 if he draws two chips with different numbers.
Since there are four chips in the box with numbers {1, 1, 3, 5}, the possible outcomes for drawing two chips are:
- (1, 1)
- (1, 3)
- (1, 5)
- (1, 3)
- (1, 5)
- (3, 5)
Out of these 6 possible outcomes:
- Only 1 of them is a win (both chips being the same) leading to a gain of \[tex]$2: (1, 1). - The remaining 5 outcomes are losses (different numbers) leading to a loss of \$[/tex]1.
Thus, the probability table is:
[tex]\[ \begin{array}{|c|c|c|} \hline X_1 & 2 & -1 \\ \hline P(X_1) & \frac{1}{6} & \frac{5}{6} \\ \hline \end{array} \][/tex]
### b. Miguel's Expected Value
The expected value [tex]\( E(X) \)[/tex] is calculated by multiplying each outcome [tex]\( X_i \)[/tex] by its probability [tex]\( P(X_i) \)[/tex] and summing these products:
[tex]\[ E(X) = X_1 \cdot P(X_1) + X_2 \cdot P(X_2) \][/tex]
Substituting the given values:
[tex]\[ E(X) = 2 \cdot \frac{1}{6} + (-1) \cdot \frac{5}{6} \][/tex]
This calculation simplifies to:
[tex]\[ E(X) = \frac{2}{6} - \frac{5}{6} = -\frac{3}{6} = -0.5 \][/tex]
Thus, the expected value [tex]\( E(X) \)[/tex] is [tex]\(-0.5\)[/tex].
### c. Expected Money Won or Lost Each Time
Based on the expected value calculated in the previous step, Miguel should expect to lose on average [tex]\(-\$0.5\)[/tex] (50 cents) each time he plays the game. This outcome means that over a large number of games, Miguel will lose 50 cents per game on average.
### d. Making the Game Fair
To make the game fair, the expected value [tex]\( E(X) \)[/tex] should be zero. Let [tex]\( Y \)[/tex] be the new value assigned for the case when Miguel wins by choosing two chips with the number 1.
The new equation for a fair game will be:
[tex]\[ Y \cdot P(X_1) + (-1) \cdot P(X_2) = 0 \][/tex]
Substituting the probabilities:
[tex]\[ Y \cdot \frac{1}{6} + (-1) \cdot \frac{5}{6} = 0 \][/tex]
Solving for [tex]\( Y \)[/tex]:
[tex]\[ \frac{Y}{6} - \frac{5}{6} = 0 \][/tex]
[tex]\[ \frac{Y}{6} = \frac{5}{6} \][/tex]
Multiplying both sides by 6:
[tex]\[ Y = 5 \][/tex]
Therefore, to make the game fair, the value that should be assigned to picking two chips with the number 1 should be [tex]\( \$5 \)[/tex].
In summary:
- The expected value is [tex]\(-0.5\)[/tex].
- Miguel should expect to lose 50 cents each game.
- To make the game fair, winning with two number 1 chips should be worth \$5.
