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

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Manager
Joined: 11 Oct 2013
Posts: 111
Concentration: Marketing, General Management
GMAT 1: 600 Q41 V31
In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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11 Dec 2015, 03:50
1
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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Manager
Joined: 31 Dec 2016
Posts: 81
In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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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

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.
Intern
Joined: 27 Sep 2017
Posts: 3
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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02 Oct 2017, 22:38
lagomez Can you explain how you got 5/6?
Manager
Joined: 10 Dec 2011
Posts: 99
Location: India
Concentration: Finance, Economics
GMAT Date: 09-28-2012
WE: Accounting (Manufacturing)
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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01 Jan 2018, 04:32
My solution but wrong

4 siblings have one sibling each (A1, A2, B1, B2) and 3 have 2 each (C1, C2, C3). Thus total possibilities of selecting 2 from 7 = 7C2 = 21

Now, when the two selected are not siblings, therefore, 1 from one group and another from the other group. Thus, selection = 4C1*3C1 = 12.

Therefore, solution = 12/21 = 4/7
Intern
Joined: 21 May 2018
Posts: 11
GMAT 1: 770 Q50 V45
Re: In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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15 Jul 2018, 00:47
Hi,

Have a quick question on whether my approach is right - I got the answer of 16/21, but just wanted to double check.

P(pair selected are siblings) = (4/7)(1/6) + (3/7)(2/6) = 10/42 = 5/21
P(pair is not siblings) = 1 - 5/21 = 16/21

(4/7) = from 7, select 1 person from the first group of 4
(1/6) = select his sibling from the rest of the group

(3/7) = from 7, select 1 person from the second group of 3
(2/6) = select either of his sibling from the rest of the group

Manager
Joined: 01 Feb 2017
Posts: 155
In a room filled with 7 people, 4 people have exactly 1  [#permalink]

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11 Aug 2018, 05:28
I have used alphabet analogy to solve this Q.

Consider three groups, each containing same alphabets.

Group 1: AA
Group 2: BB
Group 3: CCC

We need to select 2 distinct alphabets.

TOTAL OUTCOMES:
ways of selecting 2 alphabets out of 7 without any restriction is 7C2= 21

FAVORABLE OUTCOMES:
based on restriction of selecting distinct alphabets, we have following possibilities:

1 each from G1&G2: 2C1 x 2C1= 4
1 each from G1&G3: 2C1 x 3C1= 6
1 each from G2&G3: 2C1 x 3C2= 6
Total favorable outcomes: 16

Probability: 16/21

Ans E

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In a room filled with 7 people, 4 people have exactly 1 &nbs [#permalink] 11 Aug 2018, 05:28

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