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In a room filled with 7 people, 4 people have exactly 1

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Joined: 11 Oct 2013
Posts: 121

Kudos [?]: 67 [0], given: 137

Concentration: Marketing, General Management
GMAT 1: 600 Q41 V31
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In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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New post 11 Dec 2015, 03:50
reply2spg 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?

A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21


This is how I solved.

Case 1: The group of 4 have exactly 1 sibling. So maximum 2 people can be picked from that group who are not siblings.
Case 2: The group of 3 has 2 siblings each. So maximum only 1 person from the group can be chosen.
Total number of ways in which we can pick people is 7C2 = 21.


Case 1 -
\(2C1 * 2C1\) ways

Case 2 -
\(3C1 * 4C1\) ways

Probability -
\(\frac{2C1 * 2C1 + 4C1 * 3C1}{21} = \frac{16}{21}\)

Experts, please confirm if my approach is correct :)

+Kudos if this helped!
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Kudos [?]: 67 [0], given: 137

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Joined: 31 Dec 2016
Posts: 91

Kudos [?]: 11 [0], given: 22

In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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New post 12 Aug 2017, 11:49
lagomez wrote:
You have people 1-2-3-4-5-6-7

4 have exactly 1 sibling which can mean:
1-2 are siblings
3-4 are siblings

3 have exactly 2 siblings which can mean:
5-6-7 are siblings

Let's start with the group of 4:
The probability of picking 1 is (1/7)
The probability of not getting a sibling pair is (5/6) because the only other sibling is 2
Therefore, if 1 is selected first the probability of not getting a sibling pair is 5/42
Multiply that by 4 because the probability is the same whether you start with 1-2-3-4 so you get 20/42 for the first 4 people

Now let's go to the group of 3:
The probability of picking 5 is (1/7)
The probability of not getting a sibling pair is (4/6) which the non-sibling pair is 1-2-3-4
Therefore the probability is 4/42. Multiply that probability by 3, which represent 5-6-7 so the probability is 12/42

Now you have two probabilities: 12/42 and 20/42
add both and you get 32/42 or 16/21



It's easier to use the complement. Not sure why people are trying so hard here to use the choose formula when it makes the question 1000X harder

4/7 chance to pick a person with 1 sibling * 1/6 chance you get a sibling or 4/42

3/7 change to pick a 2 person sibling * 2/6 or 6/42

Total of 10/42 of getting a sibling.

We get to use the simple or rule not the generalized since the things are mutually exclusive. And we aren't drawing twice just once. So either a 4 person is picked first or 3 person.

So that simplifies to 5/21 then 1 minus this is 16/21 for the odds that they are not siblings. 5/21 they are.

Kudos [?]: 11 [0], given: 22

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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

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New post 02 Oct 2017, 22:38
lagomez Can you explain how you got 5/6?

Kudos [?]: 5 [0], given: 2

Re: In a room filled with 7 people, 4 people have exactly 1   [#permalink] 02 Oct 2017, 22:38

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In a room filled with 7 people, 4 people have exactly 1

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