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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.

2 Answers 2

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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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    (...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.2011-05-18
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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