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Probability Question: How to tackle this faster?

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Probability Question: How to tackle this faster?

by Kaunteya » Wed Feb 20, 2008 11:52 am
I have issues with probability questions. I can get to the answer but it takes me a good 3-4 minutes. Can someone please explain to me an easier method to solve this question that may save me time. I have to always draw out a picture of the people ect...wasting valuable time.

In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
a. 5/21
b. 3/7
c. 4/7
e. 5/7
f. 16/21

Merci beaucoup!
K
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by siddarthd2919 » Wed Feb 20, 2008 3:23 pm
can u post the answer.....if possible the steps.....i am stuck in this one.
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by Kaunteya » Wed Feb 20, 2008 3:41 pm
Begin by counting the number of relationships that exist among the 7 individuals whom we will call A, B, C, D, E, F, and G.

First consider the relationships of individual A: AB, AC, AD, AE, AF, AG = 6 total. Then consider the relationships of individual B without counting the relationship AB that was already counted before: BC, BD, BE, BF, BG = 5 total. Continuing this pattern, we can see that C will add an additional 4 relationships, D will add an additional 3 relationships, E will add an additional 2 relationships, and F will add 1 additional relationship. Thus, there are a total of 6 + 5 + 4 + 3 + 2 + 1 = 21 total relationships between the 7 individuals. Alternatively, this can be computed formulaically as choosing a group of 2 from 7: 7!/(2! 5!) = 21

We are told that 4 people have exactly 1 friend. This would account for 2 "friendship" relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 friends. This would account for another 3 "friendship" relationships (e.g. EF, EG, and FG). Thus, there are 5 total "friendship" relationships in the group.

The probability that any 2 individuals in the group are friends is 5/21. The probability that any 2 individuals in the group are NOT friends = 1 – 5/21 = 16/21. The correct answer is E.

To get to the answer in 2-2.5 minute range is certainly not easy. I had to think about this one for about 3-4 minutes before I tried to answer it. Even then that took me way too long. Maybe someone out there has a shortcut for solving these types of questions.

Cheers
K
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This is the only way..

by deltasquare » Thu Feb 21, 2008 11:37 pm
The way Kaunteya has done it is the best and the only way to arrive at an answer for such questions.
If we try to do it any faster than this, we are likely to make a mistake while solving.
Such problems are typical of Manhattan tests, but its unlikely that such questions will occur in GMAT test unless we've been scoring very highly (getting a lot of questions in 700+ difficulty level right)..
To make it faster we can try to eliminate obviously wrong options and guess among the likely ones.. Can't think of any other way.
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by iwillmakeit » Sat Mar 01, 2008 9:22 pm
p=1-(4/7*1/6+3/7*2/6)=16/21

4/7 - the probability to choose one of 4 people who have 1 friend.
1/6 - the probability to choose a friend of 6 remained people

3/7 - the probability to choose one of 3 people who have 2 friend.
2/6 - the probability to choose a friend of 6 remained people.
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