5 cards drawn w/o replacement: find expected number of aces

[kvm113]

Five cards are drawn from a deck without replacement. The number of aces is counted. Find the expected number of aces.

 

[DrMike]

Five cards are drawn from a deck without replacement. The number of aces is counted. Find the expected number of aces.

One really nice thing about expected values is that it doesn't matter if it's with or without replacement....

E(X+Y) = E(X) + E(Y), whether or not X and Y are independent.

So, if you can work out the expected number of aces in one card, you can just multiply that by 5.

Alternatively, note that the distribution of the number of aces follows a binomial distribution. (Can you find the parameters n and p?) Then the expected value is np