# I need some help with a probability problem, dealing with selecting cards

The Scenario is as follows:

You have three cards, A, B and C. If I pick a card four times, each time replacing the card, what are the chances I get:

4As? 3As? 2As? 1A? 0As?

I know that it has to be a total probability out of 81 (3x3x3x3), but after that I am lost.

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These are the coefficients of $(1/81)(A+2)^4$ (when you multiply it out). Can you see why?

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The answer is correct, but asking 'can you see why' does not explain why. – Kirk Broadhurst May 18 '11 at 5:05
@Kirk, if Confused wants me to explain why (or if you want me to explain why), I'll be happy to. My hope was that Confused would see why, in which case more serious learning has transpired than if I were to write out all the details. – Gerry Myerson May 18 '11 at 6:23
Yes, I'd like to you explain why. – Kirk Broadhurst May 18 '11 at 6:54
@Kirk, OK, here goes. If you multiply out $(A+B+C)^4$, you get a bunch of terms like $ABAC$ and $BBBA$ and $CAAC$, each of which corresponds to one possible outcome of picking four cards with replacement. E.g., $ABAC$ corresponds to picking $A$, then $B$, then $A$ again, and finally $C$. Now all we care about is how many $A$s. If we let $B=C=1$ (so we're expanding $(A+2)^4$), then terms like $ABAC$ with 2 $A$s become $A^2$; in general, we get $A^k$ precisely when we have $k$ $A$s. So the coefficient of $A^k$ is the number of ways of getting exactly $k$ $A$s. Now divide by $81$ (continued...) – Gerry Myerson May 18 '11 at 13:08
(...continuation) which is the total number of outcomes, to get the probabilities. OK? By the way, where is Confused? He/she/it really ought to come back here and accept an answer or else explain why no answer suits. – Gerry Myerson May 18 '11 at 13:11

You already saw that there are 81 possible ways to draw the cards (3 ways each time, 4 draws).

Now ask yourself: how many different ways are there to draw the A four times? Three times? etc

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