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sasho [114]
2 years ago
14

You roll two fair dice. If they land with a sum of 7, you get a $10. If they land with a sum of 6 or 8, you get $5. Everything e

lse earns $0. Which of the following would be the highest value you could set the price and the player would still win?
$8
$5
$2
$10​
Mathematics
1 answer:
Sindrei [870]2 years ago
3 0

Using the expected value, it is found that the highest value you could set the price is: $5.

-----------------------------

  • A game is fair, that is, the player would still win, if the <u>expected value is 0</u>.
  • The expected value is the <u>sum of each outcome multiplied by it's probability</u>.
  • A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.

  • For two fair dice, there are 36 total outcomes, as 6^2 = 36.
  • 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, \frac{6}{36} 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, \frac{8}{36} probability of getting x.
  • 36 - 14 = 22, thus \frac{22}{36} probability of losing x.

Since we want the expected value to be 0:

\frac{6}{36}(10) + \frac{8}{36}(5) - \frac{22}{36}x = 0

\frac{60 + 50 - 22x}{36} = 0

22x = 110

x = \frac{110}{22}

x = 5

The highest value you could set the price is: $5.

A similar problem is given at brainly.com/question/24905256

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