probability question

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probability question [#permalink]

29 Oct 2011, 22:08

From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the
probability that equal numbers of boys and girls will be selected?

A.1/10
B.4/9
C.1/2
D.3/5
E.2/3

OA=D.

I understand getting the numerator, (3 possibilities for males x 3 possibilities for females) and I know the answer explanation has the denominator being 6!/(4!2!) =15 but when I count out the possibilities. I only get 14 for the denominator. Where am I going wrong?

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Re: probability question [#permalink]

01 Nov 2011, 16:22

here is how i did it:
The overall number of outcomes is 6!/(4!2!). This is how we find the number of different groups of 4 using 6 people. 6!/(4!2!) = 15

We can make 3 different groups of 2 girls using 3 girls. The same goes for the boys, 3 groups as well. 3x3= 9, we multiply because one group of girls can be paired with 3 boys groups.

Thus, 9/15 = 3/5 D

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Re: probability question [#permalink]

02 Nov 2011, 11:13

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Stringworm, your list of possibilities is complete--it just doesn't add up to 14 because it's the answer to a different question. You have shown all the ways to put 2 indistinguishable boys and 2 indistinguishable girls in order. What we want for our denominator is a list of all the ways we can choose 4 particular (distinguishable) children out of 3 boys and 3 girls. We aren't concerned with what order the children appear in, but we are concerned with which children are selected. The list looks like this:

(g1)(g2)(b1)(b2)
(g1)(g2)(b1)(b3)
(g1)(g2)(b2)(b3)
(g1)(g3)(b1)(b2)
(g1)(g3)(b1)(b3)
(g1)(g3)(b2)(b3)
(g2)(g3)(b1)(b2)
(g2)(g3)(b1)(b3)
(g2)(g3)(b2)(b3)
(b1)(all 3 girls)
(b2)(all 3 girls)
(b3)(all 3 girls)
(g1)(all 3 boys)
(g2)(all 3 boys)
(g3)(all 3 boys)

Note that the first 9 choices are the ones we want in the numerator. Also note that we would not want to spend time doing this on the test. If you can't find a faster way to do it, drop the problem!
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Re: probability question [#permalink]

09 Nov 2011, 19:24

DmitryFarber, Thank you very much. That makes sense.

Re: probability question [#permalink] 09 Nov 2011, 19:24

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