John has coins in a jar. Five of the coins are pennies, 8 are dimes, and 3 are quarters. John randomly selects a coin from the jar and without replacing it
randomly selects another coin. What is the probability that he selects a dime and
then a quarter?

Step-by-step explanation:

X = 10 → 2 nickels

P(1st Nickel) = 5/16

P(2nd Nickel) = 4/15

P(X = 10) = P(2 nickels) = (5/16)*(4/15) = 20/240 = 0.08333 (8.333%)

The probability of selecting 2 nickels ($0.10) is 0.08333.

X = 11 → dime and penny

P(Dime) = 3/16

P(Penny) = 8/15

P(X = 11) = P(Dime and Penny) = (3/16)*(8/15) = 24/240 = 0.1 (10%)

You could also choose penny first and then a dime. This will give you the same probability.

P(Penny) = 8/16

P(Dime) = 3/15

P(X = 11) = P(Penny and Dime) = (8/16)*(3/15) = 24/240 = 0.1 (10%)

P(Dime and Penny) + P(Penny and Dime) = 0.2 (20%)

The probability of selecting a dime and a penny no matter the order is 0.2.

Total Value = 8(0.01) + 5(0.05) + 3(0.10) = 0.08 + 0.25 + 0.30 = $0.63

There are a total of 16 coins and you selected 2 coins from the jar. So the probability of selecting 2 coins from the jar is 2/16 or 0.125.

E[X] = ($0.63)*(2/16) = $0.07875

The expected value is about 8 cents.

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