SOLUTION: 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 $200; if she loses, she must pay $18. If X = am

Question 867113: 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 $200; if she loses, she must pay $18. If X = amount of money Mary wins (or loses), what is the expected value of X? (Round your answer to the nearest cent.)

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
E[X] = (Probability of Winning)*(Value of Winning) + (Probability of Losing)*(Value of Losing)

E[X] = (0.02)*(200) + (0.98)*(-18)

E[X] = (4) + (-17.64)

E[X] = -13.64

Her expected value is -13.64 dollars

So she expects to lose $13.64 every time she plays (this is on average).
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