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There are 60 balls in an urn. Ten balls each are marked each letter A through F. Six balls are selected without replacement. What is the probability that the balls selected are one of each letter A through F? Order does not matter.

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What have you tried? Are you stuck at any particular point? –  David Mitra Feb 6 '12 at 3:00
    
All choices are equally likely. I believe the denominator is $\binom{60}{6}$, but I do not know about the numerator. –  jamil Feb 6 '12 at 3:12
    
This should help: en.wikipedia.org/wiki/… –  Byron Schmuland Feb 6 '12 at 3:18

2 Answers 2

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Hint:

(i) How many ways are there to choose $6$ balls from $60$? Make sure that the choices are all equally likely.

(ii) How many ways are there to select $1$ ball from the $10$ with label A, and $1$ from the $10$ with label B, and so on up to $1$ from the $10$ with label F?

In solving this problem, imagine that the balls are all distinct, that they are people, $10$ wearing a red shirt, $10$ wearing a blue shirt, and so on.

Another way: Pick the balls one at a time. First ball doesn't matter. What is the probability that the second ball is of a different colour than the first? Now given that your first two balls were of different colours, what is the probability that the third ball is of a new colour? So what is the probability that the first three balls are of different colours?

Now given that on your first three picks your got all different colours, what is the probability that the fourth pick brings us a new colour? Continue.

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Thank you for pointing out and; I originally and incorrectly tried to use addition of the $\binom{10}{1}$ instead of multiplication –  jamil Feb 6 '12 at 3:22

$$\frac{\binom{10}{6}}{\binom{60}{6}}=\frac{3}{59\times58\times57\times11}$$ This comes from the fact that total number of outcomes are ${\binom{60}{6}}$, whereas number of favourable outcomes are $\binom{10}{6}$.

Update: I misinterpreted the "each" in the question. Here's the new answer: $$\frac{10^6}{\binom{60}{6}}=\frac{1,000,000}{50,063,860}=0.1997$$ Imagine that each set of A-F balls has a color. Now, you can choose an A ball in 10 ways, a B ball in 10 ways, and so on. So $10^6$ is the total number of ways of choosing one of each letter. The denominator is pretty obvious, its the number of ways of choosing 6 balls out of 60 distinct balls (We have forcibly made them distinct by painting them).

In any probability problem, you can always "paint" identical items (or number them if they already have a color). As long as you take this change into account in both your numerator and denominator, you should be fine. This works as it does not matter to us in the end if the balls that we chose had a color or not; if you think about it, the probability of getting all balls A-F would not change if you painted them. Think about it (Imagine actually doing this experiment and measuring probability; then painting the balls and doing it all over again).

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Maybe it's too late for me, but I don't see how your numerator is correct. There are 10 ways to select the "A" ball, 10 ways to select the "B" ball, ... What you have, I believe, is the probability that all the balls are "A" balls. –  David Mitra Feb 6 '12 at 3:12
    
Oh, so that's what you meant by ten balls "each". Sorry, my bad. I misinterpreted the question. –  Manishearth Feb 6 '12 at 3:16
    
I updated the answer. –  Manishearth Feb 6 '12 at 3:25

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