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I hate Probability questions..plz help

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I hate Probability questions..plz help

by sana.noor » Sat Feb 02, 2013 10:17 am
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

E is the right choice, I know that their are 21 different ways to choose 2 people but How to find 16 is really tough for me.
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by Brent@GMATPrepNow » Sat Feb 02, 2013 10:24 am
sana.noor 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
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by GMATGuruNY » Sat Feb 02, 2013 10:33 am
I posted an explanation here:

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by ziko » Mon Feb 04, 2013 11:48 pm
If there are 3 people out of 7 who have exactly 3 sublings than there should be 3 sublings (whithin that group of 7), and 4 people who have exactly 1 subling there must be two pairs of sublings.
SO, for me it is much easier when i allocate so letters, let say there are AAABBCC. So the quastion asks for the probability of chosing two people so there will not be any sublings.
Option 1)
First step - The probability to take one person from A's is 3/7 multiplied by 4/6 (the probability to chose the second person from other sublings, so for us it does not matter which one to choose as long as he/she is not from A's)
Second step - the probability to take one person from B's 2/7 multiplied by probability of chosing a second person not from B's, 5/6. We need to multiply this result by 2 because we have two such scenarios, namely with B' and C's.
Final step - Add all the probability cases, because we can have either of them, 12/42+20/42=32/42=16/21

I hope that this explanation hepls!
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