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Try this one on for size - courtesy of Manhattan GMAT

Expert replies
by pkw209 » Tue Feb 02, 2010 5:15 pm
Question:

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

Answer to come...
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Source: — Problem Solving |

by fibbonnaci » Tue Feb 02, 2010 8:12 pm
this is a simple one provided u get the logic rite.

lets say A, B, C, D are the 4 ppl who have one sibling. and E, F, and G are the people who have 2 siblings.

eg say A has a sibling within the 4 people and E has a sibling within the three people.

so in order for selecting 2 people in such a way that they are not siblings, we must select people from different groups.

selecting one person from the 4 membered group is 4C1= 4
selecting one person from the 3 membered group is 3C1 = 3

number of ways= 12 (4*3)

total number of possibilities of selecting 2 people out of the 7 membered group= 7C2= 21

probability= 12/21 => 3/7

the probability of selecting 2 individuals such that they are not siblings is 3/7

Hope this helps!
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by Stuart@KaplanGMAT » Tue Feb 02, 2010 11:15 pm
fibbonnaci wrote:this is a simple one provided u get the logic rite.

lets say A, B, C, D are the 4 ppl who have one sibling. and E, F, and G are the people who have 2 siblings.

eg say A has a sibling within the 4 people and E has a sibling within the three people.

so in order for selecting 2 people in such a way that they are not siblings, we must select people from different groups.

selecting one person from the 4 membered group is 4C1= 4
selecting one person from the 3 membered group is 3C1 = 3

number of ways= 12 (4*3)

total number of possibilities of selecting 2 people out of the 7 membered group= 7C2= 21

probability= 12/21 => 3/7

the probability of selecting 2 individuals such that they are not siblings is 3/7

Hope this helps!
Not quite!

You've assumed that you have to select one person from each group. However, in the first group of 4, each person only has 1 sibling, i.e. there are 2 pairs of siblings in that set of 4 people. So, there are some pairs you can pick out of that group of 4 who won't be siblings.

We could use your approach, but we have to break it down further:

Siblings: AB, CD, EFG

We can pick one person from any two of these three groups and end up with a non-sibling pair.

Case 1: 1 from AB, 1 from CD: 2*2 = 4 pairs
Case 2: 1 from AB, 1 from EFG: 2*3 = 6 pairs
Case 3: 1 from CD, 1 from EFG: 2*3 = 6 pairs

So, we can make 16 non-sibling pairs.

There are 7C2 = 21 total pairs available.

Accordingly, the probability of a non-sibling pair is 16/21.
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by ajith » Wed Feb 03, 2010 3:27 am
Stuart Kovinsky wrote:
Siblings: AB, CD, EFG

We can pick one person from any two of these three groups and end up with a non-sibling pair.

Case 1: 1 from AB, 1 from CD: 2*2 = 4 pairs
Case 2: 1 from AB, 1 from EFG: 2*3 = 6 pairs
Case 3: 1 from CD, 1 from EFG: 2*3 = 6 pairs

So, we can make 16 non-sibling pairs.

There are 7C2 = 21 total pairs available.

Accordingly, the probability of a non-sibling pair is 16/21.
Wouldn't it be easier to find out the probability of the pair being siblings and deduct it from 1?

The pair possibilities are ( AB, CD, EF, EG and FG) in total 5 out of 21
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by pkw209 » Wed Feb 03, 2010 2:06 pm
Stuart is correct! 3/7 is the trap answer if the question isn't fully understood.

Ajith's approach is also effective.
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by tnaim » Thu Jun 17, 2010 6:22 am
I am having a hard time understanding how "4 people have exactly 1 sibling in the room" translates to just 2 siblings relationships. Let's say A,B,C, and D each represent a person, and let's say B is the sibling; from that I thought we'd have:
AB (sibling relationship)
BC(sibling relationship)
BD(sibling relationship)
With this logic, I end up getting 5/7 as the answer, which is wrong, but I cannot figure out what it is that I am doing wrong.
Thank you!!
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by Testluv » Thu Jun 17, 2010 7:02 am
tnaim wrote:I am having a hard time understanding how "4 people have exactly 1 sibling in the room" translates to just 2 siblings relationships. Let's say A,B,C, and D each represent a person, and let's say B is the sibling; from that I thought we'd have:
AB (sibling relationship)
BC(sibling relationship)
BD(sibling relationship)
With this logic, I end up getting 5/7 as the answer, which is wrong, but I cannot figure out what it is that I am doing wrong.
Thank you!!
See if this helps: https://www.beatthegmat.com/probability-t59130.html
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by Stuart@KaplanGMAT » Thu Jun 17, 2010 11:06 am
tnaim wrote:I am having a hard time understanding how "4 people have exactly 1 sibling in the room" translates to just 2 siblings relationships. Let's say A,B,C, and D each represent a person, and let's say B is the sibling; from that I thought we'd have:
AB (sibling relationship)
BC(sibling relationship)
BD(sibling relationship)
With this logic, I end up getting 5/7 as the answer, which is wrong, but I cannot figure out what it is that I am doing wrong.
Thank you!!
Hi,

according to your arrangement, A, B, C and D are all siblings ("sibling" means brother/sister - so by definition, if Alan and Betty are siblings, and Betty and Charles are siblings, Alan and Charles are also siblings). This violates the "have exactly 1 sibling in the room" condition imposed by the question stem.

The only way to conform to the rule is:

AB are siblings
CD are siblings

(A has exactly 1 sibling: B;
B has exactly 1 sibling: A;
C has exactly 1 sibling: D; and
D has exactly 1 sibling: C).
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