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In a room filled with 7 people, 4 people have exactly 1 sibling in the
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Author:  Bunuel [ 08 Oct 2014, 03:43 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

keyrun wrote:
Can anyone please help me on where I went wrong?


When you select groups 1 and 2, you have 3*2 choices, when groups 1 and 3 you have 3*2 choices and when groups 2 and 3 you have 2*2 choices: total = 6 + 6 + 4.

Author:  BrentGMATPrepNow [ 13 May 2019, 10:36 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

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


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

Let's use counting techniques to answer this question

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

Answer: E

Cheers,
Brent

Author:  ScottTargetTestPrep [ 18 Sep 2019, 08:54 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

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


Let A, B, C, D, E, F, and G be the 7 people in the room. To satisfy the condition that 4 people have exactly 1 sibling and 3 people have exactly 2 siblings, we can let A and B be siblings (but not to other people), C and D be siblings (but not to other people), and E, F and G are siblings (but not to other people).

Let’s consider the probability of how each person is chosen:

If A is chosen first, then B can’t be chosen. So the probability is:

1/7 x 5/6 = 5/42

This probability will be the same if B, C, or D is chosen first.

If E is chosen first, then neither F nor G can be chosen. So the probability is:

1/7 x 4/6 = 4/42

This probability will be the same if F or G is chosen first.

Therefore, the overall probability is:

5/42 x 4 + 4/42 x 3 = 20/42 + 12/42 = 32/42 = 16/21

Alternate Solution:

Notice that 2 people can be chosen out of 7 people in 7C2 = 7!/(5!*2!) = (7 x 6)/2 = 21 ways.

With A, B, C, D, E, F, and G as above, we see that there are 5 ways to choose a sibling pair: A-B, C-D, E-F, E-G and F-G. Thus, 21 - 5 = 16 choices of do not include a sibling pair. Therefore, the probability that the chosen two people are not siblings is 16/21.

Answer: E

Author:  BrentGMATPrepNow [ 09 Dec 2020, 06:12 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

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


First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio, which we'll call "the TRIPLETS"
- a sibling pair, which we'll call "the TWINS"
- and another sibling pair, which we'll also call "the TWINS"

So, P(two are NOT siblings) = P(1st choice is a triplet AND 2nd choice is not a sibling OR 1st choice is a twin AND 2nd choice is not a sibling)
= P(1st choice is a triplet) x P(2nd choice is not a sibling) + P(1st choice is a twin) x P(2nd choice is not a sibling)
= (3/7 x 4/6) + (4/7 x 5/6)
= 12/42 + 20/42
= 32/42
= 16/21

Answer: E

Author:  Bambi2021 [ 05 Apr 2021, 01:38 ]
Post subject:  In a room filled with 7 people, 4 people have exactly 1 sibling in the

Three groups can be combined in pairs of two groups A, B, C in AB + AC + BC = 3 ways where

A = 2 people
B = 2 people
C = 3 people

AB = 2C1*2C1 = 4 different ways to chose two people
AC = 2C1*3C1 = 6 different ways
BC = 2C1*3C1 = 6 different ways

Total number of ways to chose any two people out of seven = 7C2 = 21 ways

16/21

Author:  rvgmat12 [ 04 Sep 2021, 06:17 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

3 people in one group (each has two siblings)

4 people ( 2 groups of siblings )

Total combinations of 2 people selected not being siblings = 1 - (3C2+2)/7C2 = 1-5/21 = 16/21

E

Author:  M838TE [ 31 Mar 2022, 11:35 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

IanStewart
Hi, ian.
Would you mind explaining approach 3 that bunuel accounted for?
I am not sure if I understand the multiplication by 2 for the 2-sibling group.
Does order matter here? Not sure why picking 1-2, 3-4 then the 5-6-7 are different than picking 5-6-7 first, then [1-2], [3-4].

Understanding the concept by picking one person at a time was easier for me: 1/7 to pick any*5/6 because one person is picked and the other person in the group is a sibling that is picked up. The numerator is the total ways of picking a non-sibling pair, order does not matter and position does not matter, then you multiply by 4 because there are 4 diff ppl with 1-sibling to pick from.

Posted from my mobile device

Author:  IanStewart [ 31 Mar 2022, 11:53 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

M838TE wrote:
IanStewart
Hi, ian.
Would you mind explaining approach 3 that bunuel accounted for?


Bunuel explained that solution a bit further down on page one of this thread. He first divided the seven people into sibling groups, which is a complication there's no need to think about here. We have a 4/7 probability of picking someone with one sibling first, then a 1/6 chance of picking their only sibling second. We have a 3/7 probability of picking someone with two siblings first, then a 2/6 chance of picking one of their two siblings second. So the probability we get a pair of siblings is (4/7)(1/6) + (3/7)(2/6) = 5/21, and the probability we don't is 1 - (5/21) = 16/21.

You could also use this approach, but solve directly, by finding the probability right away that you do not pick a sibling on the second selection (then you get (4/7)(5/6) + (3/7)(4/6) = 16/21 and do not need to subtract the result from 1 at the end). Bunuel's method #3 does essentially that, though it considers more cases, because it treats each sibling group separately.

Author:  M838TE [ 31 Mar 2022, 14:41 ]
Post subject:  In a room filled with 7 people, 4 people have exactly 1 sibling in the

IanStewart wrote:
M838TE wrote:
IanStewart
Hi, ian.
Would you mind explaining approach 3 that bunuel accounted for?


Bunuel explained that solution a bit further down on page one of this thread. He first divided the seven people into sibling groups, which is a complication there's no need to think about here. We have a 4/7 probability of picking someone with one sibling first, then a 1/6 chance of picking their only sibling second. We have a 3/7 probability of picking someone with two siblings first, then a 2/6 chance of picking one of their two siblings second. So the probability we get a pair of siblings is (4/7)(1/6) + (3/7)(2/6) = 5/21, and the probability we don't is 1 - (5/21) = 16/21.

You could also use this approach, but solve directly, by finding the probability right away that you do not pick a sibling on the second selection (then you get (4/7)(5/6) + (3/7)(4/6) = 16/21 and do not need to subtract the result from 1 at the end). Bunuel's method #3 does essentially that, though it considers more cases, because it treats each sibling group separately.

I actually saw his explanation. But I was confused by the part where he multiply by2.
specifically this part, " Multiplying by 2 since this selection can be don in two ways: the first from {5, 6, 7} and the second from {1, 2} or {3, 4} OR the first from {1, 2} or {3, 4} and the second from {5, 6, 7};"
Posted from my mobile device

Author:  JJDa [ 19 Jul 2022, 15:11 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

I found this nice explanation on youtube

https://www.youtube.com/watch?v=Lefh_Ywk3qw

Author:  ThatDudeKnows [ 20 Jul 2022, 06:05 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

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


We have three groups: AA, BB, CCC
We can "win" in two ways:
1. We select either A or B and then select someone other than their sibling. \(\frac{4}{7}*\frac{5}{6} = \frac{20}{42}\)
OR
2. We select C and then someone other than their sibling. \(\frac{3}{7}*\frac{4}{6} = \frac{12}{42}\)
\(\frac{20}{42}+\frac{12}{42} = \frac{32}{42} = \frac{16}{21}\)
Answer choice E.

We could also have found the probability of "losing" and subtracted from 1.
1. We select either A or B and then select their sibling. \(\frac{4}{7}*\frac{1}{6} = \frac{4}{42}\)
OR
2. We select C and then their sibling. \(\frac{3}{7}*\frac{2}{6} = \frac{6}{42}\)
\(\frac{4}{42}+\frac{6}{42} = \frac{10}{42} = \frac{5}{21}\)
\(1-\frac{5}{21} = \frac{16}{21}\)
Answer choice E.

Author:  BIGDAMNGOD [ 01 Oct 2022, 08:43 ]
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the

I think I have a conceptual gap here, Experts please help -

My approach -

Group A - 2 siblings (4 people)
Group B - 3 siblings (3 people)

4C1* 5C1 = 20
3C1* 4C1 = 12

Total = 7C2 = 21

Why does the total selection 7C2 not work this way?

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