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sibling probability

Expert replies
by dikku07 » Mon Oct 05, 2009 1:38 pm
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
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Source: — Problem Solving |

Re: sibling probability

by Stuart@KaplanGMAT » Mon Oct 05, 2009 2:31 pm
dikku07 wrote: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
As with most GMAT questions, there are a number of different approaches we can use to solve. Let's look at one of them.

First, let's talk formulas:

Probability = # desired outcomes / total # of possibilities

is the basic probability formula. Chances are pretty good that you use that one already. However, there's another key formula that's a huge time saver on the GMAT:

Prob(want) = 1 - Prob(don't want)

There are a lot of GMAT probability questions on which it's quicker to calculate the probability of what you don't want to happen and subtract that from 1.

On this question, we could certainly take that approach and set up the formula:

Prob(not siblings) = 1 - Prob(are siblings)

Prob(are siblings) = (# of sibling pairs we could possibly pick)/(total # of possible pairs).

Based on the original info, we know that 4 of them have exactly 1 sibling and 3 have exactly 2 siblings. The only way that can be broken down with 7 people is:

1 pair of sibs
1 pair of sibs
1 trio of sibs

So, how many sibling pairs can we make?

From each pair of sibs we can, not surprisingly, make 1 pair of sibs... so that's 2.

From the trio of sibs, we can make 3C2 ("three choose two" is how we pronounce that in the combinatorics world) = 3!/2!1! = 3 pairs of sibs. Of course, you really don't need a fancy formula to figure that out - you can just think "If I have 3 people A, B and C, how many pairs can I make? AB/AC/BC... that's 3 pairs."

So, we have 5 different sibling pairs.

For the total number of possible pairs, we do want to use the combinations formula:

7C2 = 7!/2!5! = 7*6/2 = 21

Therefore, the probability of choosing a pair of siblings is: 5/21

Is that one of the answers? Of course it is. The GMAT knows that a large number of test takers will be so happy that they completed a calculation that they'll forget what the original question was! For just that reason, we have Step 4 of the Kaplan Method for Problem Solving: double check the question.

We're careful, so we realize that the correct answer to the original question isn't 5/21, it's:

1 - 5/21 = 16/21... choose (E).
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by dikku07 » Mon Oct 05, 2009 2:59 pm
Thanks Stuart for in depth explanation. thats correct answer
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