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

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Joined: 02 Sep 2009
Posts: 60645
Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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30 Jun 2019, 10:10
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:
$$P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}$$.

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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18 Sep 2019, 09:54
2
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.

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

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23 Dec 2019, 08:06
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:
$$P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}$$.

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 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
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10011
Location: Pune, India
Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the  [#permalink]

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26 Dec 2019, 01:00
1
varotkorn wrote:
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

Yes, this solution is correct, no problem.
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Karishma
Veritas Prep GMAT Instructor

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the   [#permalink] 26 Dec 2019, 01:00

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