No Complementary Method Please

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No Complementary Method Please

by knight247 » Thu Oct 06, 2011 7:18 am
A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left- and right-hand glove of the same color) will be among the three gloves selected?

(A) 3/10
(B) 23/60
(C) 7/12
(D) 41/60
(E) 5/6

The OA is D. Its fairly simple when u find the probability of not getting a single pair. But hoping to get a direct solution without using the complementary method. Detailed explanations would be appreciated.
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by GmatMathPro » Thu Oct 06, 2011 9:09 am
Here is one way:

You will either get a blue or green glove on your first draw. Chance of blue=3/5, green=2/5.

If you draw blue first:

blue first,Blue match,any glove=(3/5)(3/9)(1)=1/5
blue first,any non-matching glove, blue match=(3/5)(6/9)(3/8)=3/20
blue first, green, green matching=(3/5)(4/9)(2/8)=1/15

green first:

green first, green matching, any glove=(2/5)(2/9)(1)=4/45
green first, any non matching glove, green match=(2/5)(7/9)(2/8)=7/90
green first, blue, matching blue=(2/5)(6/9)(3/8)=1/10

Add them all up to get [spoiler]41/60[/spoiler]
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