Find the expected value (or expectation) of the games described.
A bag contains eight white balls and two black balls, John picks two balls at random from the bag, and he wins $\$5$ if he does not pick a black ball.
The bag has 10 balls consisting of eight white balls and two black balls. The probability of getting a white is $\displaystyle \frac{8}{1} = \frac{4}{5}$ and the probability of getting a black ball is $\displaystyle \frac{2}{10} = \frac{1}{5}$
Assuming that the ball is not replaced after the first one has been picked, the probability that he picks first white ball is $\displaystyle \frac{8}{10}$ and the probability that he picks second white ball is $\displaystyle \frac{7}{9}$. So the expected winning value is
$\displaystyle 5 \left( \frac{8}{10} \right)\left( \frac{7}{9} \right) = \frac{28}{9} = 3.11$
This means that if you play this game, you will make, on average $\$3.11$ per game.
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