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Sagot :
Using the expected value, it is found that you should charge $4 to have a fair game.
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- A probability is the number of desired outcomes divided by the number of total outcomes.
- The expected value is the sum of each outcome multiplied by it's respective probability.
- When you roll two dice, there are [tex]6^2 = 36[/tex] possible outcomes.
- We suppose a cost of x.
- In 6 of them, the sum is 7: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1).
- Thus, [tex]\frac{6}{36}[/tex] probability of earning a profit of 12 - x.
- 2 of them have sums of 2 or 12, which are (1,1) and (6,6).
- Then, [tex]\frac{2}{36}[/tex] probability of earning a profit of 36 - x.
- If the outcome is any of the remaining 28, the player loses, thus [tex]\frac{28}{36}[/tex] probability of losing x.
- The game is fair if the expected value is 0, then:
[tex]\frac{6}{36}(12 - x) + \frac{2}{36}(36 - x) - \frac{28}{36}x = 0[/tex]
[tex]\frac{6(12 - x) + 2(36 - x) - 28x}{36} = 0[/tex]
[tex]72 + 72 - 36x = 0[/tex]
[tex]36x = 144[/tex]
[tex]x = \frac{144}{36}[/tex]
[tex]x = 4[/tex]
A value of $4 should be charged.
A similar problem is given at https://brainly.com/question/24855677
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