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Probability qt

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Probability qt

by akshaykerur » Sat Dec 10, 2011 2:34 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?
5/21
3/7
4/7
5/7
16/21

Can someone please solve this for me in detail. MGMAT didnt have a very good explanation.

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by pemdas » Sat Dec 10, 2011 5:21 pm
since 4 people having 1 sibling and 3 people having 2 siblings considered separately would make 4*(1+1)+3*(1+2)=17 which is greater than 7, we are looking for situation when 4 people and 3 people are siblings with each other in their groups.

The ONLY possible scenario within the group of 7 people -> 4 people each is sibling of the other (each has one sibling) + 3 people each is sibling of the other two people (each has two siblings).

4 people ABCD each is sibling of the other -> we can select AB or CD or AD ... 4C2 or 6 ways are present if we happen to select among 4 out of 7 people, (4/7)*(1/6)
3 people EFG each has two siblings -> we can select EFG or FGE or GEF ... 3C2 or 3 ways are present if we happen to select among 3 out 7 people, (3/7)*(1/3)
(4/7)*(1/6)+(3/7)*(1/3)=2/21 +1/7=5/21

The possibility we don't select two people who are siblings 1-5/21=16/21
akshaykerur 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

Can someone please solve this for me in detail. MGMAT didnt have a very good explanation.

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by Anurag@Gurome » Sun Dec 11, 2011 8:16 pm
akshaykerur 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

Can someone please solve this for me in detail. MGMAT didnt have a very good explanation.

Source: MGMAT CAT
Probability that the two individuals chosen are NOT siblings = 1 - probability that the two individuals chosen are siblings
Let the 7 people are A, B, C, D, E, F, G.
Now, 4 people have exactly 1 sibling implies let A, B, C, and D have exactly 1 sibling. So, A-B are siblings and C-D are siblings.
Next, 3 people have exactly 2 siblings implies let E has F and G, both as his siblings. This means that F has E and G as siblings, and G has E-F as siblings.
So, the groups of siblings are: A-B, C-D, E-F-G
Now, 2 individuals are selected from the room at random, so ways of choosing these 2 people = 7C2 = 21 ways
No. of ways of choosing a sibling pair = 1 + 1 + 3 = 5 ways
Probability that the two individuals chosen are siblings = 5/21
Therefore, probability that the two individuals chosen are NOT siblings = 1 - 5/21 = [spoiler]16/21[/spoiler]

The correct answer is E.
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by akshaykerur » Sun Dec 11, 2011 9:08 pm
Thank you very much Anurag.
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by Anurag@Gurome » Sun Dec 11, 2011 9:13 pm
akshaykerur wrote:Thank you very much Anurag.
My pleasure, Akshay.
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by pemdas » Sun Dec 11, 2011 9:49 pm
what, if the question changed ...
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 three individuals are selected from the room at random, what is the probability that there are no siblings?
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by pemdas » Mon Dec 12, 2011 3:28 am
let me post my way of solving this changed q.

Since the groups of siblings are: A-B, C-D, E-F-G or A-C, B-D, E-F-G or A-F, C-E, B-D-G, etc. we may deduce

The variability within the two-people groups is 4C2 or 6 pairs. Only two pairs out of six are having siblings of each other, the other four pairs have not siblings.
Each of the two pairs (having siblings) can be attached three more people from the three-people group (having siblings), this makes triplets 2*3 of three-people groups (having siblings).
Because we have one triplet of people (having siblings) among total possible 7C3 (or 35) triplets too, we account for 1 more three-people group. Total makes (6+1) three-people groups/triplets of siblings out of possible 35 triplets -> 7/35=1/5

Probability of selecting three people having no siblings is 1 - 1/5 = 4/5
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by BN89 » Mon Dec 12, 2011 9:21 am
Anurag@Gurome wrote:
akshaykerur 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

Can someone please solve this for me in detail. MGMAT didnt have a very good explanation.

Source: MGMAT CAT
Probability that the two individuals chosen are NOT siblings = 1 - probability that the two individuals chosen are siblings
Let the 7 people are A, B, C, D, E, F, G.
Now, 4 people have exactly 1 sibling implies let A, B, C, and D have exactly 1 sibling. So, A-B are siblings and C-D are siblings.
Next, 3 people have exactly 2 siblings implies let E has F and G, both as his siblings. This means that F has E and G as siblings, and G has E-F as siblings.
So, the groups of siblings are: A-B, C-D, E-F-G
Now, 2 individuals are selected from the room at random, so ways of choosing these 2 people = 7C2 = 21 ways
No. of ways of choosing a sibling pair = 1 + 1 + 3 = 5 ways
Probability that the two individuals chosen are siblings = 5/21
Therefore, probability that the two individuals chosen are NOT siblings = 1 - 5/21 = [spoiler]16/21[/spoiler]

The correct answer is E.
Great answer, but could you please explain why there's 5 ways to choose a sibling pair?

edit: nevermind, already figured it out.
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