Question

A jar contains 3 pennies, 7 nickels and 2 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.

Find the probability X = 10.    

Find the probability X = 11.    

Find the expected value of X.    

Answer 1

It is given that a jar contains 3 pennies, 7 nickels and 2 dimes. A child selects 2 coins at random without replacement from the jar. Let X represent the amount in cents of the selected coins.

So, the total number of coins = 3+7+2 = 12

First, we have to find P(X = 10). To get 10 cents by selecting two coins from the jar, the coins must be nickels. Thus, the probability that the first coin is a nickel = 7/12. The probability that the second coin is a nickel = 6/11.

Thus, the probability that the two coins are nickel = (7/12)*(6/11) = (7*6)/(12*11) = 42/132

Thus, P(X = 10) = 42/132

Now, we have to find P(X = 11). To get 11 cents by selecting two coins from the jar, the coins must be a dime and a penny. Thus, the probability that the first coin is a dime and the second penny = (2/12)*(3/11). The probability that the first coin is a penny and the second dime = (3/12)*(2/11).

Thus, the probability that the two coins are dime and penny = {(2/12)*(3/11)}+{(3/12)*(2/11)} = (6/132)+(6/132) = 12/132

Thus, P(X = 11) = 12/132

Now, we have to find the expected value of X. For this, we have to find the other probabilities.

P(X = 2) : This can be done when both the coins are pennies. Thus, the probability of first coin to be a penny = 3/12. The probability of the second coin to be a penny = 2/11. Thus, the probability that the two coins are pennies = (3/12)*(2/11) = 6/132.

Thus, P(X = 2) = 6/132

P(X = 6) : This can be done when one coin is a penny and the other is a nickel. Thus, the probability of the first coin to be a penny and the second nickel = (3/12)*(7/11) = 21/132. The probability of the first coin to be a nickel and the second penny = (7/12)*(3/11) = 21/132. Thus, the probability that the coins are nickel and penny = (21/132)+(21/132) = 42/132

Thus, P(X = 6) = 42/132

P(X = 15) : This can be done when one coin is a nickel and the other is a dime. Thus, the probability that the first coin is a nickel and the second dime = (7/12)*(2/11) = 14/132. The probability that the first coin is a dime and the second nickel = (2/12)*(7/11) = 14/132. Thus, the probability that the coins are a nickel and dime = (14/132)+(14/132) = 28/132

Thus, P(X = 15) = 28/132

P(X = 20) : This can be done when both the coins are dimes. Thus, the probability that the first coin is a dime = 2/12. The probability that the second coin is a dime = 1/11. Thus, the probability that the two coins are dimes = (2/12)*(1/11) = 2/132

Thus, P(X = 20) = 2/132

Thus, Expected value of X = X*P(X) = 2(6/132) + 6(42/132) + 10(42/132) + 11(12/132) + 15(28/132) + 20(2/132) = 12/132 + 252/132 + 420/132 + 132/132 + 420/132 + 40/132 = (12+252+420+132+420+40)/132 = 1276/132 = 9.67

Thus, Expected value of X = 9.67 .