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Tough Probability Problem

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Tough Probability Problem

by razorback » Thu Oct 06, 2011 1:44 pm
From MGMAT... Their explanation really isnt clicking with me. Please take a shot!

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21
B) 3/7
C) 4/7
D) 5/7
E) 16/21

OA: E

BTW: The first thing I did was compute the total number of possible outcomes, which I do fortunately understand. Two random selections from 7, and their order doesn't matter = YYNNNNN = 7!/5!2! = 21. After that I might as well have been reading French!
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by GmatMathPro » Thu Oct 06, 2011 1:58 pm
The question implies that there are two pairs of siblings and one group of 3 people who are all siblings. Let's represent it as follows: A-B C-D E-F-G.

So A,B,C,D are the people who have exactly one sibling, and E,F,G are the people who have exactly 2 siblings.

You correctly calculated that there are 21 ways to choose two people from the room. Notice that there are exactly 5 ways to get a pair of siblings: you could pick A-B, C-D, E-F, F-G, or E-G. So there is a 5/21 chance of getting a pair of siblings. We want the probability of NOT getting siblings, so subtract this value from 1 to get [spoiler]16/21[/spoiler]
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by razorback » Thu Oct 06, 2011 2:11 pm
Oh DUH! Thanks!
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by GMATGuruNY » Thu Oct 06, 2011 7:36 pm
I posted a solution here:

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