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A game of chance is played using a six-sided die. If a 1,2 or 3 is rolled the player wins $0 if a 4 or 5 is rolled the player wins $7. If a 6 is rolled the player wins $10. If each play of the game costs 5$ what is the expected pay off?

Respuesta :

CuteIzzy

Answer:

-2.50

Step-by-step explanation:

Using the formula for the expected value, it is found that the expected pay off is of -$1.

  • The expected value is given by the sum of each outcome multiplied by it's respective probability.

In this problem, the outcomes and probabilities are, considering the cost of the game of $5 and that there are 6 sides:

  • [tex]\frac{3}{6}[/tex] probability of losing $5.
  • [tex]\frac{2}{6}[/tex] probability of winning $2.
  • [tex]\frac{1}{6}[/tex] probability of winning $5.

Hence:

[tex]E(X) = \frac{3}{6}(-5) + \frac{2}{6}(2) + \frac{1}{6}(5) = \frac{3(-5) + 2(2) + 1(5)}{6} = -\frac{6}{6} = -1[/tex]

The expected pay off is of -$1.

To learn more about expected value, you can take a look at https://brainly.com/question/24855677

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