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A person must pay $ $ 8 to play a certain game at the casino. Each player has a probability of 0.21 of winning $ $ 14, for a net gain of $ $ 6 (the net gain is the amount won 14 minus the cost of playing 8). Each player has a probability of 0.79 of losing the game, for a net loss of $ $ 8 (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 probabiltiy distribution)? If the Expected Value is negative, be sure to include the "-" sign with the answer. Express the answer with two decimal places.

Answer :

Answer:

-$5.06

Step-by-step explanation:

Given the probability distribution of X where X is the net gain or loss

[tex]\left|\begin{array}{c|c|c}Profit(X)&\$6&-\$8\\P(X)&0.21&0.79\end{array}\right|[/tex]

The expected value of X is defined as follows:

Expected Value of X, [tex]E(X)=\sum_{i=1}^nx_iP(x_i)[/tex]

Therefore, the expected value of the player

E(X)=(6*0.21)+(-8*0.79)

=1.26-6.32

[tex]E(X)=-\$5.06[/tex]

The expected value of each player at the casino is -$5.06.

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