A jar contains 5 pennies, 5 nickels and 8 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.

Question

A. Find the probability X = 10.

B. Find the probability X = 11

C. Find the expected value of X. 



Answer 1

X = 10 → 2 nickels

P(1st Nickel) = 5/18

P(2nd Nickel) = 4/17

P(X = 10) = P(2 nickels) = (5/18)*(4/17) = 20/306 = 0.06536 (6.536%)

The probability of selecting 2 nickels ($0.10) is 0.06536.

X = 11 → dime and penny

P(Dime) = 8/18

P(Penny) = 5/17

P(X = 11) = P(Dime and Penny) = (8/18)*(5/17) = 40/306 = 0.13072 (13.072%)

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

P(Penny) = 5/18

P(Dime) = 8/17

P(X = 11) = P(Penny and Dime) = (5/18)*(8/17) = 40/306 = 0.13072 (13.072%)

P(Dime and Penny) + P(Penny and Dime) = 0.26144 (26.144%)

The probability of selecting a dime and a penny no matter the order is 0.26144.

Total Value = 5(0.01) + 5(0.05) + 8(0.10) = 0.05 + 0.25 + 0.80 = $1.10

There are a total of 18 coins and you selected 2 coins from the jar. So the probability of selecting 2 coins from the jar is 2/18 or 0.1111.

E[X] = ($1.10)*(2/18) = $0.1222

The expected value is about 12 cents.