Question

A bowl contains 10 chips, of which 8 are marked $$\$ 2$$ each and 2 are marked $$\$ 5$$ each. Let a person choose, at random and without replacement, three chips from this bowl. If the person is to receive the sum of the resulting amounts, find his expectation.

Short answer

The expected amount of money that the person will receive is \$6.8

Step-by-step solution

List all possible outcomes

The person can pick 3 chips from the bowl resulting in the amounts: \$6, \$7, \$8, \$9 and \$12. Corresponding to the combinations (2,2,2), (2,2,5), (2,5,2), (5,2,2) and (5,5,2). Note that (5,5,2) can only occur if the \$2 chip is picked last, while other outcomes can occur in multiple ways.

Calculate the probability for each outcome

Next, calculate the probabilities for each outcome:The probability for \$6 is \(\frac{8}{10} * \frac{7}{9} * \frac{6}{8} = \frac{28}{45}\) as there are three \$2 chips chosen from 8, then 7, then 6 (as the chips aren't replaced).The probability for \$7 is \(\frac{8}{10} * \frac{7}{9} * \frac{2}{8} * 3 = \frac{28}{45}\), the 3 accounts for the three different orders the chips can be chosen in.The probability for \$9 is \(\frac{8}{10} * \frac{2}{9} * \frac{1}{8} * 3 = \frac{1}{15}\), the 3 accounts for the three different orders the chips can be chosen in.The probability for \$12 is \(\frac{2}{10} * \frac{1}{9} * \frac{8}{8} = \frac{2}{45}\), this only occurs in one order.

Calculate the expectation

Lastly, multiply each outcome by its probability and sum these to get the expectation: \((\$6 * \frac{28}{45}) + (\$7 * \frac{28}{45}) + (\$9 * \frac{1}{15}) + (\$12 * \frac{2}{45}) = \$6.8\).