Question

The probability that Mary will win a game is 0.02, so the probability that she will not win is 0.98. If Mary wins, she will be given $160; if she loses, she must pay $16. If X = amount of money Mary wins (or loses), what is the expected value of X? (Round your answer to the nearest cent.)

Answer 1

Given:

Probability of winning, P(X) = 0.02

Probability of losing, P($\bar{X}$) = 0.98

Wining amount = $160

Losing amount = $16

Step-by-step explanation:

Let the expected amount of money win be  'X'

Expected value of X, E(X) = Probability of winning, P(X).Probability of winning, P(X)  - Probability of losing, P($\bar{X}$).Losing amount

Now,

E(X) = ($0.02\times 160 - 0.98\times 16$)

E(X) = -12.48

Expected value of X = -12.48

Expected loss value = $12.48 loss