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There are 6 of x and 4 of y. A group of 5 is selected. What is the probability of there being 3x and 2 y? I completely forgot how to do this and my textbook doesn't have any information on it.

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    @Sivaram: An "answer" here doesn't have to be a complete answer. That is, if you want to avoid giving a complete answer to homework and just give a hint, it's perfectly fine to post the hint as an "answer." Questions with no upvoted answers are considered "unanswered" and periodically get bumped back to the top of the front page.2011-01-19

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We have $6$ $x$'s and $4$ $y$'s. Hence, the total number of $x$'s and $y$'s is $10$.

The total number of ways $5$ of them are selected is $C(10,5)$.

The number of ways of choosing $3$ $x$'s from a pool of $6$ $x$'s is $C(6,3)$. The number of ways of choosing $2$ $y$'s from a pool of $4$ $y$'s is $C(4,2)$.

So the total number of favorable cases is $C(6,3) \times C(4,2)$.

Hence, the probability is $\frac{C(6,3) \times C(4,2)}{C(10,5)} = \frac{10}{21}$