Using the expected value, it is found that the highest value you could set the price is: $5.
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- A game is fair, that is, the player would still win, if the expected value is 0.
- The expected value is the sum of each outcome multiplied by it's probability.
- A probability is the number of desired outcomes divided by the number of total outcomes.
For two fair dice, there are 36 total outcomes, as [tex]6^2 = 36[/tex].
The charge is x.
6 outcomes result in a sum of 7, which are (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1), thus, [tex]\frac{6}{36}[/tex] probability of getting 10.
10 outcomes result in a sum of 6 or 8, which are (1,5), (2,4), (2,6), (3,3), (3,5), (4,2), (4,4), (5,1), (5,3) and (6,2), thus, [tex]\frac{8}{36}[/tex] probability of getting x.
36 - 14 = 22, thus [tex]\frac{22}{36}[/tex] probability of losing x.
Since we want the expected value to be 0:
[tex]\frac{6}{36}(10) + \frac{8}{36}(5) - \frac{22}{36}x = 0[/tex]
[tex]\frac{60 + 50 - 22x}{36} = 0[/tex]
[tex]22x = 110[/tex]
[tex]x = \frac{110}{22}[/tex]
[tex]x = 5[/tex]
The highest value you could set the price is: $5.
A similar problem is given at https://brainly.com/question/24905256
