Question 1161812: A person must pay $9 to play a certain game at the casino. Each player has a probability of 0.11 of winning $15, for a net gain of
$6 (the net gain is the amount won 15 minus the cost of playing 9).
Each player has a probability of 0.89 of losing the game, for a net loss of
$9 (the net loss is simply the cost of playing since nothing else is lost).
What is the Expected Value for the player (that is, the mean of the probability distribution)? If the Expected Value is negative, be sure to include the "-" sign with the answer. Express the answer with two decimal places.
Expected Value = $
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
probability of winning = 0.11
probability of losing = 0.89
note that the two probabilities add to 1.00, which represents 100%
Expected value = (probability of winning)*(net value of winning) + (probability of losing)*(net value of losing)
Expected value = (0.11)*(6) + (0.89)*(-9)
Expected value = -7.35
Answer: -7.35 dollars
The player expects on average to lose $7.35 each time the game is played
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