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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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@a sandwhich: Have you heard of permutations and combinations (remember the 'C' and 'P' notations?)? It will be of help in this problem. Count the total number of favorable ways and the total number of ways(using the 'C' 's). The ratio will give you the probability. –  user17762 Jan 18 '11 at 23:15
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Yes, I have. So would it be 6C3 * 4C2 / 10C5? –  a sandwhich Jan 18 '11 at 23:19
    
Yes precisely... –  user17762 Jan 18 '11 at 23:27
    
@Sivaram: It would be helpful if you repost the content of your original comment as an answer, so that this question could become "answered." –  Isaac Jan 19 '11 at 1:56
    
@Isaac: I was wondering if I could write out a solution to the homework problem. But anyway a sandwhich figured it out himself. So I guess there is nothing wrong in writing out the answer. –  user17762 Jan 19 '11 at 2:34

1 Answer 1

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

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