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You play a gambling game with a friend in which you roll a die. If a 1 or 2 comes up, you win $8. Otherwise you lose $2. What is your expected value for this game? Round your answer to the nearest cent.

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thaovtp1407

Answer:

The expected value $[tex]\frac{4}{3}[/tex]  

Step-by-step explanation:

Given:

  • If a 1 or 2 comes you win $8.
  • If a 3, 4, 5 ,6 comes, you loose $2

So:

  • The probability you win : [tex]\frac{2}{6}[/tex]
  • The probability you loose: [tex]\frac{4}{6}[/tex]

Hence, the expected value  for this game is the sum of products of probability and value:

= [tex]\frac{2}{6}[/tex] *$8 + [tex]\frac{4}{6}[/tex] *$2

= $[tex]\frac{4}{3}[/tex]

Hope it wil find you well.

The expected value for this game 1.33 dollars.

Important information:

  • A die is rolled.
  • If a 1 or 2 comes up, you win $8. Otherwise, you lose $2.

We need to find the expected value for this game.

Expected value:

If a die is rolled then the possible outcomes are 1, 2, 3, 4, 5, 6.

1, 2 means you win and 3, 4, 5, 6 means you lose.

If a 1 or 2 comes up, you win $8. Otherwise, you lose $2. So, the expected value for this game is:

[tex]E(x)=8\times P(\text{Win})-2\times P(\text{Lose})[/tex]

[tex]E(x)=8\times \dfrac{2}{6}-2\times \dfrac{4}{6}[/tex]

[tex]E(x)=8\times \dfrac{1}{3}-2\times \dfrac{2}{3}[/tex]

[tex]E(x)=\dfrac{8}{3}-\dfrac{4}{3}[/tex]

[tex]E(x)=\dfrac{4}{3}[/tex]

[tex]E(x)\approx 1.33[/tex]

Thus, the expected value for this game 1.33 dollars.

Find out more about 'Expected value' here:

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