College Algebra, Chapter 10, 10.5, Section 10.5, Problem 12

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.