Stuggling with Problem

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avada: Senior | Next Rank: 100 Posts; Posts: 38; Joined: Fri Aug 17, 2012 11:28 am; Thanked: 2 times

by avada » Fri Aug 17, 2012 11:35 am

This problem is taken from the Manhattan online Cat practice exams. I understand how to get the proper number of relationships(ie 5 present) but am struggling to understand the total combinations possible. There should be 7*6 combinations but I am not sure about the rest. Thank you

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?
5/21
3/7
4/7
5/7
16/21

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Source: Beat The GMAT — Problem Solving | Fri Aug 17, 2012 11:35 am

rijul007: Legendary Member; Posts: 588; Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members; Location: New Delhi, India; Thanked: 130 times; GMAT Score:720

by rijul007 » Fri Aug 17, 2012 12:10 pm

avada 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?
5/21
3/7
4/7
5/7
16/21

Lets say the 7 people are
A B C D E F G

and
A and B are siblings
C and D are siblings
E, F and G are siblings

Total no of pairs that can be selected = 7C2 = 6*7/2 = 21
Probability that those two individuals are siblings = P
Probability that those two individuals are NOT siblings = 1 - P

from A, B, C and D, exactly 2 pairs can be made, such that, the two individuals are siblings
from E, F and G, exactly 3C2 pairs can be made

Total no of pairs in which two individuals are siblings = 2 + 3C2 = 2 + 3 = 5

P = 5/21
Probability that those two individuals are NOT siblings = 1 - P = 1 - 5/21 = 16/21

Option E

avada wrote:This problem is taken from the Manhattan online Cat practice exams. I understand how to get the proper number of relationships(ie 5 present) but am struggling to understand the total combinations possible. There should be 7*6 combinations but I am not sure about the rest. Thank you

7*6 combinations would include (A,B) as well as (B,A). Both of them are the same, so in order to avoid repetition we divide 7*6 by 2.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Fri Aug 17, 2012 1:13 pm

avada 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?
5/21
3/7
4/7
5/7
16/21

First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio
- a sibling pair
- and another sibling pair

Using counting techniques:

For this question, it's easier to find the complement.
So P(not siblings) = 1 - P(they are siblings)

P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]

# of ways to select 2 siblings
Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)

So, total number of ways to select 2 siblings = 3+1+1 = 5

total # of ways to select 2 people
We have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)

So, P(they are siblings) = 5/21

This means P(not siblings) = 1 - 5/21
= 16/21 = E

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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