Join IDNLearn.com today and start getting the answers you've been searching for. Our platform provides trustworthy answers to help you make informed decisions quickly and easily.
Sagot :
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.
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.
