Question

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?

Answer 1

Answer:

-$13.5

Step-by-step 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