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The probability that mary will win a game is 0.04, so the probability that she will not win is 0.96. if mary wins, she will be given $200; if she loses, she must pay $14. if x = amount of money mary wins (or loses), what is the expected value of x? (round your answer to the nearest cent.)

Respuesta :

calculista
Let
EV----------> expected value

we know that

EV = (Probability of Winning)*(Value of Winning) + (Probability of Losing)*(Value of Losing)
so
EV=0.04*(200)+0.96*(-14)-----> EV=8+(-13.44)-----> EV=-$5.44
So
she expects to lose $5.44 every time she plays (this is on average)

the answer is
-$5.44

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