43.6k views
3 votes
A game is played using ne die. If the die is rolled and shows 3, the player wins $45. If the die shows any number other than 3, the player wins nothing. If there is a charge of $9 to play the game, what is the game's expected value?

User Caseynolan
by
8.7k points

1 Answer

6 votes

Answer:

-$13.5

Explanation:

Let x be a random variable of a count of player gain.

- We are told that if the die shows 3, the player wins $45.

- there is a charge of $9 to play the game

If he wins, he gains; 45 - 9 = $36

If he looses, he has a net gain which is a loss = -$9

Thus, the x-values are; (36, -9)

Probability of getting a 3 which is a win is P(X) = 1/6 since there are 6 numbers on the dice and probability of getting any other number is P(X) = 5/6

Thus;

E(X) = Σ(x•P(X)) = (1/6)(36) + (5/6)(-9)

E(X) = (1/6)(36 - (5 × 9))

E(X) = (1/6)(36 - 45)

E(X) = -9/6 = -3/2

E(X) = -3/2

This represents -3/2 of $9 = -(3/2) × 9 = - 27/2 = -$13.5

User Krn
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories