SOLUTION: I am really struggling with this problem. I know the first is 4/52 or 0.769 or 76.9% but the rest I am really struggling with: From a single deck of cards, you select a card, loo

Question 590785: I am really struggling with this problem. I know the first is 4/52 or 0.769 or 76.9% but the rest I am really struggling with:
From a single deck of cards, you select a card, look at it, put it back in the deck, and shuffle. If you do this 10 times, what is the probability of drawing exactly (1) one ace, (b) two aces, (c) three aces, (d) four aces (each of these is calculated separately)? Out of the 10 draws of cards, what is the mean number of aces you would expect to find?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
From a single deck of cards, you select a card, look at it, put it back in the deck, and shuffle. If you do this 10 times, what is the probability of drawing exactly (1) one ace, (b) two aces, (c) three aces, (d) four aces (each of these is calculated separately)? Out of the 10 draws of cards, what is the mean number of aces you would expect to find?
----
Replacement implies Binomial.
n = 10 and P(ace) = 1/13
-----
(1) one ace,; P(x = 1) = 10C1*(1/13)(12/13)^9 = 0.3743
(b) two aces,; P(x = 2) = 10C2*(1/13)^2(12/13)^8 = 0.1404
(c) three aces, ; p(x = 3) = 0.0312
(d) four aces ; p(x = 4) = 0.0045
----------------------
(each of these is calculated separately)? Out of the 10 draws of cards, what is the mean number of aces you would expect to find?
E(x) = np = 10*(1/13) = 0.7692
================
Cheers,
Stan H.
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