SOLUTION: A jar contains 2 pennies, 6 nickels and 4 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 1134926: A jar contains 2 pennies, 6 nickels and 4 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 = 11.

Find the expected value of X
x=10 is 5/22 but i cannot figure out the rest
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

If you were able to determine that P(X=10) is 5/22, then you should be able to work the whole problem.

P(X=10) is the probability that the value of the two coins is 10 cents. That means the two coins were both nickels. So in choosing 2 of the 12 coins, he must select 2 of the 6 nickels. The probability is

That can of course be simplified to 5/22.

But let's not do that yet. To find the expected value for the problem, we need to find the probabilities of all the possible combinations of coins. We can verify that our calculations are correct by seeing that the sum of all the probabilities is 1. So leaving all of the probabilities in unsimplified form (with denominator C(12,2)=66) makes sense.

I'll do a couple of the other cases for you; you can do the remaining cases, including the one that the problem asks for -- P(X=11).

P(X=2). The two coins have to be both of the 2 pennies. The probability is

P(X=15). The two coins have to be 1 of the 4 dimes AND 1 of the 6 nickels. The probability is

Do the other cases (X=6, X=11, and X=20) and make sure the probabilities sum to 66/66 = 1.

Then I will assume you know how to find the expected value once you have the probabilities. In cents, it will be:

(2 times P(X=2)) + (6 times P(X=6)) + ... + (20 times P(X=20))

(NOTE: The answer for the expected value, surprisingly, is a whole number....)
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