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

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

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New post 03 Jun 2019, 12:56
I calculated the probability they would be siblings:

(4/7 *1/6) + (3/7 * 2/6) = 10/42 = 5/21 and then i subtracted this from 1 to get 16/21.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 30 Jun 2019, 06:27
Bunuel 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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Solution #1:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

\(P=1-\frac{5}{21}=\frac{16}{21}\).

Solution #3:
\(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\).

Answer: E.




Bunuel

in solution 1, why are we taking it to 3 sections, after combination from both the groups i.e \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings),
why cant we select one from triple and then any one from the rest for as they wont be siblings to each other.

So i took it like
selections of 2 people which are not siblings = \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings) + \(C^1_3*C^1_4\) (one from triple*one from rest)

What is wrong in here?? I'm really confused.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 30 Jun 2019, 09:10
Sreeragc wrote:
Bunuel 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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Solution #1:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

\(P=1-\frac{5}{21}=\frac{16}{21}\).

Solution #3:
\(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\).

Answer: E.




Bunuel

in solution 1, why are we taking it to 3 sections, after combination from both the groups i.e \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings),
why cant we select one from triple and then any one from the rest for as they wont be siblings to each other.

So i took it like
selections of 2 people which are not siblings = \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings) + \(C^1_3*C^1_4\) (one from triple*one from rest)

What is wrong in here?? I'm really confused.


I think I already answered similar doubt here: https://gmatclub.com/forum/in-a-room-fi ... l#p1100701
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In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 23 Dec 2019, 07:06
Bunuel wrote:
Sreeragc wrote:
Bunuel 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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4).
As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Solution #1:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2:
# of selections of 2 out of 7 - \(C^2_7=21\);
# of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

\(P=1-\frac{5}{21}=\frac{16}{21}\).

Solution #3:
\(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\).

Answer: E.




Bunuel

in solution 1, why are we taking it to 3 sections, after combination from both the groups i.e \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings),
why cant we select one from triple and then any one from the rest for as they wont be siblings to each other.

So i took it like
selections of 2 people which are not siblings = \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings) + \(C^1_3*C^1_4\) (one from triple*one from rest)

What is wrong in here?? I'm really confused.


I think I already answered similar doubt here: https://gmatclub.com/forum/in-a-room-fi ... l#p1100701


Dear Bunuel VeritasKarishma IanStewart,

I think Sreeragc 's counting approach is also correct:
\(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings) = 4
\(C^1_3*C^1_4\) (one from triple*one from rest) = 12

The total is 16, which is the same as Bunuel 's Solution #1 in the first page.
There is no double-counting problem as suggested in the link posted by Bunuel (https://gmatclub.com/forum/in-a-room-fi ... l#p1100701)

Is this a valid way?,
Thank you
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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New post 01 Feb 2020, 21:46
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 are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.

Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.

The correct answer is E.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the   [#permalink] 01 Feb 2020, 21:46

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