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

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Re: Probability - Siblings in the room [#permalink]

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23 Jun 2011, 11:02
cant we solve this by this method given in a Gclub test ???
M#06-10
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?

soln:
once we select the first person, then there is 1/11 chance that the second person selected is their spouse and 10/11 chance that the second person selected is not their spouse. For the third person, there is a 8/10 chance that the third is not a spouse of the first two, and for the fourth person, there is a 6/9 probability that the fourth is not married to any of the first three.

hence :
$\left(\frac{10}{11}\right)*\left(\frac{8}{10}\right)*\left(\frac{6}{9}\right)=\frac{16}{33}$

can someone map this method to the problem in the thread ?

This GMAT club solution is much better. I tried this in the MGMAT cat problem, got sucked and wasted 4 minutes to no avail.
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Re: Probability - Siblings in the room [#permalink]

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23 Jun 2011, 11:26
bblast wrote:
cant we solve this by this method given in a Gclub test ???
M#06-10
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?

soln:
once we select the first person, then there is 1/11 chance that the second person selected is their spouse and 10/11 chance that the second person selected is not their spouse. For the third person, there is a 8/10 chance that the third is not a spouse of the first two, and for the fourth person, there is a 6/9 probability that the fourth is not married to any of the first three.

hence :
$\left(\frac{10}{11}\right)*\left(\frac{8}{10}\right)*\left(\frac{6}{9}\right)=\frac{16}{33}$

can someone map this method to the problem in the thread ?

This GMAT club solution is much better. I tried this in the MGMAT cat problem, got sucked and wasted 4 minutes to no avail.

The best best for this problem is to assume the 6 couples
A a
B b
C c
D d
E e
F f

now to select one person is 12/12 ( suppose u choose A now u cant take a)
next person will be in 10/11 ways ( suppose u choose B .. now u cant take b.. a is still left but u cant take him/her as well)
next person in 8/10 ways ( suppose u choose C , we are left with a,b, and C whom we cant choose)
4th person in 6/9 ways

total is 12/12*10/11*8/10*6/9 = 46/99 = 16/33...
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Re: Probability - Siblings in the room [#permalink]

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23 Jun 2011, 21:51
another way to solve this would be as follows

we got 3 groups of siblings

AB

CD

EFG

we have to select 2 from these and they cannot be siblings. so you can pick as follows

possible outcomes = one from group1,one from group2 or
one from group2 ,one from group 3 or
one from group3,one from group 1

= (2*2 +2*3+3*2)/(7c2) = 16/21 .

alternative approach

you can find out ways of select siblings and deduct that from 1 . its the same.

here we need to write down the possible siblings - AB, CD, EF,GH,HE

probability of selecting two siblings= select one sibling pair = 5c1/7c2 = 5/21

probability of selecting two non siblings = 1- 5/21 =16/21

siddhans wrote:
agdimple333 wrote:
Other way - Probability that 2 individuals chosen are siblings-
from 3 groups , A- B, C-D, E-F-G
2 * 1 / 7 * 6 + 2 * 1 / 7 * 6 + 3 * 2/ 7 * 6 = 10 / 7 * 6
= 5/21
Are not siblings = 1-5/21
= 16/21

Hi sudhir,

Can you explain this method too?
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Re: Probability - Siblings in the room [#permalink]

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24 Jun 2011, 02:06
hi sudhir, appreciate the help, but My request was not to solve the gmat club problem again(its already solved very clearly in my post); but to solve the main problem of this thread using the method used.
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Re: Probability - Siblings in the room [#permalink]

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02 Oct 2011, 11:14
I have a slightly different approach. While all approaches here are great, I just feel more intuitive and comfortable with mine.

Now, the way the question is phrased, I could immediately simplify the set to AABBCCC, where AA and BB are 2 sets of 1 sibling each, and CCC is 1 set of 2 siblings.

P(no siblings) = 1 - p(siblings)

P(siblings) is basically you select A and A or B and B or C and C. Translating this to math, we get:

2/7*1/6 + 2/7*1/6 + 3/7*2/6 = 5/21.

Therefore, the probability of no siblings is 16/21
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Re: Probability - Siblings in the room [#permalink]

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05 Dec 2011, 21:00
Expert's post
4 people have exactly 1 sibling in the room
Translate: #1 is siblings with #2. #3 is siblings with #4. Or you could mix the numbers around, but however way you arrange them, there are TWO pairs of siblings in this group.

3 people have exactly 2 siblings in the room
Translate: #1 is siblings with #2 and #3 - they're all siblings with each other. So this is ONE triplet of siblings.

So in total, you have one pair, one pair, and then one triplet.

You see the word "NOT" in the question - so you can consider doing the popular (1-prob that they ARE siblings) method.

prob (2 picked people are siblings) = ?

Well, 2 people picked can only siblings if they belong in the same pair or same triplet. We already divided up the people in the the pair, pair, and triplet.

So prob (2 picked people are siblings) = (2 choose 2) + (2 choose 2) + (3 choose 2)
Total # of possibilities = (7 choose 2)

In each case you are choosing 2 people. But your broke down the problem from a large set of 7 people to smaller groups of size 4 and size 3.

So 2nCr2 + 2nCr2 + 3nCr2 = 1 + 1 + 3 = 5 ways to pick siblings

Total # of possibilities = 7nCr2 =

7*6* (5!)
---------- = 42/2 = 21 possibilities
2 * (5!)

So # of ways to pick siblings is 5 out of a total of 21.

We want the opposite of this so we do 1 - (5/21) =16/21

Having gone through this, note that of all the answer choices - (D) is the last one I would have picked. Usually, with a question like this, if (5/21) is the correct answer, then whatever 1 - (5/21) is will surely also be an answer choice to trick people who forgot to do the 1 - operation.

Looking at the answer choices, I see that (b) 3/7 and (c) 4/7 add up to 1. So those two can potentially be correct.

(A) 5/21 and (e) 16/21 add up to 1, so they can potentially be correct.

Then (d) just comes out of nowhere as 5/7 by itself.

If I were to guess, I'd probably guess between A, B, C, and E and would eliminate D from guessing.

Just our 2 cents. Hope that helps!

http://www.gmatpill.com/practice-questions-explanations-gmat-prep/room-filled-7-people-4-people-1-sibling/
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Re: Probability - Siblings in the room [#permalink]

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05 Dec 2011, 22:45
Expert's post
bblast wrote:
cant we solve this by this method given in a Gclub test ???
M#06-10
If 4 people are selected from a group of 6 married couples, what is the probability that none of them would be married to each other?

soln:
once we select the first person, then there is 1/11 chance that the second person selected is their spouse and 10/11 chance that the second person selected is not their spouse. For the third person, there is a 8/10 chance that the third is not a spouse of the first two, and for the fourth person, there is a 6/9 probability that the fourth is not married to any of the first three.

hence :
$\left(\frac{10}{11}\right)*\left(\frac{8}{10}\right)*\left(\frac{6}{9}\right)=\frac{16}{33}$

can someone map this method to the problem in the thread ?

This GMAT club solution is much better. I tried this in the MGMAT cat problem, got sucked and wasted 4 minutes to no avail.

I don't know whether you are still looking for the MGMAT problem's solution in GMAT club solution format (since you put up this post in June), but I will anyway put it down.

First of all, notice that the two questions are different in that the groups in MGMAT question are not similar.

GMAT Club problem: Select 4 people out of 6 married couples. (similar groups - all couples)
MGMAT problem: Select 2 people out of 2 pairs of siblings and a triplet. (different groups - 2 pairs and a triplet)

How this affects our solution: We select the first person. Now the probability that the second person is not a sibling depends on the first person - whether he belonged to one of the pairs or the triplet.

Say the siblings are: AB, CD and EFG
The two people can be selected in 2 ways: Both from the pairs or One from pair, one from triplet

Probability of selecting both from the pairs: 4/7 (select any one of A, B, C or D) * 2/6 (Depending on your previous selection, select from the other pair) = 4/21

Probability of selecting from the triplet and from a pair = (3/7) (select one from the triplet) * (4/6) (select one from the pairs) * 2! (you could first select from the pair and then from the triplet too) = 4/7

Total probability = 4/21 + 4/7 = 16/21
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Re: Probability - Siblings in the room [#permalink]

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13 Dec 2011, 05:51
Based on the question, siblings in the room are something like this:
AA BB CCC

So, to calculate probability that 2 selected people must not be sibling:

$$P=\frac{2}{7}*\frac{5}{6}+\frac{2}{7}*\frac{5}{6}+\frac{3}{7}*\frac{4}{6}$$

$$P=\frac{16}{21}$$

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Re: Probability - Siblings in the room [#permalink]

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24 Sep 2013, 21:43
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Re: In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

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25 Sep 2013, 01:42
Expert's post
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 (4-5-6).

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:
$$P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html
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In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

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10 Aug 2014, 21:53
hussi9 wrote:
agdimple333 wrote:
Other way - Probability that 2 individuals chosen are siblings-
from 3 groups , A- B, C-D, E-F-G
2 * 1 / 7 * 6 + 2 * 1 / 7 * 6 + 3 * 2/ 7 * 6 = 10 / 7 * 6
= 5/21
Are not siblings = 1-5/21
= 16/21

Lets put the equaltion step wise

Probability that 2 individuals chosen are siblings-
from 3 groups , A- B, C-D, E-F-G
(2 / 7) * (1/6) + (2 / 7) * (1/6) + (3/7) * (2/6) = 5/21

Are not siblings = 1-5/21
= 16/21

I think the above is a little compressed. Let me see if the following expands it:

P(A-B or B-A) = 1/7*1/6 + 1/7*1/6 = 1/21
P(C-D or D-C) = same logic as above = 1/21
P(E-F or E-G or E-F) = 1/7*1/6+1/7*1/6+1/7*1/6 = 3/21
P(Sib) = 1/21+1/21+3/21 =5/21
P(No Sib) = 1-5/21 = 16/21
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Re: In a room filled with 7 people, 4 people have exactly 1 sibl [#permalink]

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13 Aug 2014, 00:41
Total ways of selecting individuals from group is C(7,2)
Let us split the group into a set of 3 (each with 2 siblings) and 2(each with 1 sibling) and 2(each with 1 sibling)

So, ways of selecting two individuals who are siblings will be = Either both individuals from the set of 3 with 2 siblings each, or both from the 1st set of 1 sibling each, or both from the 2nd set of 1 sibling each

This is equivalent to C(3,2) + C(2,2) + C(2,2)

therefore, probability of selecting a group of only siblings = C(3,2) + C(2,2) + C(2,2) / C(7,2) = 5/21

hence, probability of selecting group with no siblings = 1 - 5/21 = 16/21

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

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18 Aug 2014, 10:55
If 4 people have exactly 1 sibling, there are 2 pairs of siblings in the 4 people

Say A B C and D, each have exactly 1 sibling. That implies, A has 1 sibling (say B), so B's sibling is A
Similarly, C and D are siblings

Now if 3 people (say E F and G) each have 2 siblings, then E must have F and G both as siblings
F must have E and G both as siblings, and same for G too.

Now ways in which two ppl selected are not siblings.

1. 1 from the group of ABCD and 1 from the group of DEF. Hence number of ways = 4c1*3c1
2. _ , _ can also be filled using just ABCD in 2 * 2 ways. Each position can have 2 candidates. Hence 4 ways

Total 16 ways in which the two guys will not be siblings.

Total number of ways to select 2 ppl from 7 is 7c2 = 21

Hence reqd prob = 16/21
Re: In a room filled with 7 people, 4 people have exactly 1 sibl   [#permalink] 18 Aug 2014, 10:55

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