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M06-10

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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?

A. \(\frac{1}{33}\)
B. \(\frac{2}{33}\)
C. \(\frac{1}{3}\)
D. \(\frac{16}{33}\)
E. \(\frac{11}{12}\)
[Reveal] Spoiler: OA

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Official Solution:

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?

A. \(\frac{1}{33}\)
B. \(\frac{2}{33}\)
C. \(\frac{1}{3}\)
D. \(\frac{16}{33}\)
E. \(\frac{11}{12}\)


The probability is \(\frac{16}{33}\). Once we select the first person, then there is \(\frac{1}{11}\) chance that the second person selected is their spouse and \(\frac{10}{11}\) chance that the second person selected is not their spouse. For the third person, there is a \(\frac{8}{10}\) chance that the third is not a spouse of the first two, and for the fourth person, there is a \(\frac{6}{9}\) probability that the fourth is not married to any of the first three.

\((\frac{10}{11})*(\frac{8}{10})*(\frac{6}{9})=\frac{16}{33}\)

Alternative explanation:

\(\frac{2*2*2*2*C_6^4}{C_{12}^4}=\frac{16}{33}\)

\(C_6^4\) - represents four couples from the given 6.

Basically, we have to take 4 couples out of 6, which can be done in \(C_6^4\) ways, and then from each couple you take one person, which can be done in 2 ways for each couple. With 4 couples, this is \(2^4=2*2*2*2\) ways.

One more approach:

Say we have couples like this (A-man, B-woman):

A1 and B1

A2 and B2

A3 and B3

A4 and B4

A5 and B5

A6 and B6

So we have 6 men (6 x A) and 6 women (6 x B)

You can pick 4 people in the following ways:
  1. 4 men - i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
  2. 3 men and 1 woman - i.e. \(C_6^3 * C_3^1\) (1 of 3 and not of 6 b/c only 3 women are left to avoid 'coupling' chosen men) = \(\frac{6!}{3!*3!} * \frac{3!}{1!*2!} = 60\)
  3. 2 men and 2 women - i.e. \(C_6^2 * C_4^2\) (again 4 women left to avoid 'coupling') = \(\frac{6!}{2!*4!} * \frac{4!}{2!*2!} = 90\)
  4. 1 man and 3 women - i.e. \(C_6^1 * C_5^3 = \frac{6!}{1!*5!} * \frac{5!}{2!*3!} = 60\)
  5. 4 women - i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)

So there are \(15+60+90+60+15=240\) possibilities.

Divide this by general possibilities of picking 4 people out of 12 - i.e. \(C_{12}^4 = \frac{12!}{4!*8!} = 495\).

\(\frac{240}{495} = \frac{16}{33}\).

Yet another approach:

Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send that one "representatives" to the committee in \(C^4_6\) # of ways.

But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\).

So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\).

Total # of ways to choose 4 people out of 12 is \(C^4_{12}\).

\(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\)


Answer: D
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Re: M06-10 [#permalink]

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New post 25 Oct 2015, 09:27
I was thinking to solve it in the following way that is not correct. Please do let me know what is wrong in this approach.

Prob: No. of ways to select 4 persons who are not married to each other / No. of ways to select 4 people out of 12.

No. of ways to select 4 persons who are not married to each other: 12 * 10 * 8 * 6 (For the first person to be selected there are 12 options, once he is selected then for the second person there are 10 options and so on)

No. of ways to select 4 people out of 12 = (12 * 11 * 10 * 9) / (4 * 3 * 2) = 11 * 5 * 9

Now, when I divide (12 * 10 * 8 * 6) and (11 * 5 * 9) I don't get the answer. Could you please let me know what is wrong in my approach.

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Re: M06-10 [#permalink]

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Hi,
The top 12*10*8*6 is not selection it is arrangement.
Bottom should also be arrangement then that is 12!/8!

Else make top selection 12*10*8*6/4!
And bottom selection 12!/4!*8!

Imagine four letters abcd isn't it 4*3*2*1 arrangement. If it is 4*3*2*1/4! It is selection.
Hope I am clear

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Re: M06-10 [#permalink]

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New post 27 Oct 2015, 06:26
Hi,
The top 12*10*8*6 is not selection it is arrangement.
Bottom should also be arrangement then that is 12!/8!

Else make top selection 12*10*8*6/4!
And bottom selection 12!/4!*8!

Imagine four letters abcd isn't it 4*3*2*1 arrangement. If it is 4*3*2*1/4! It is selection.
Hope I am clear

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Re: M06-10 [#permalink]

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Hi Bunuel

I solved it through a different approach please let me know if there is any flaw in this approach


Total outcomes , say B = 12C4

Now first person can be selected from any 12 people, the next guy would have to be selected then from 10 people (leaving the person who is married to the first person selected), similarly third person can be selected from 8 and fourth person from 6

So 12*10*8*6

Now, as per question all people should be treated as same

So (12*10*8*6) should be divided by 4! (I.e.Total Arrangement / identical items arrangement)

Say A= (12*10*8*6)/4!


Now for probability divide A by B and you get 16/33.

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Re: M06-10 [#permalink]

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New post 02 Nov 2015, 10:59
Thanks a lot jojojoseph for pointing out my mistake.

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Re: M06-10 [#permalink]

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New post 20 Jun 2017, 13:10
Bunuel wrote:
Official Solution:

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?

A. \(\frac{1}{33}\)
B. \(\frac{2}{33}\)
C. \(\frac{1}{3}\)
D. \(\frac{16}{33}\)
E. \(\frac{11}{12}\)


The probability is \(\frac{16}{33}\). Once we select the first person, then there is \(\frac{1}{11}\) chance that the second person selected is their spouse and \(\frac{10}{11}\) chance that the second person selected is not their spouse. For the third person, there is a \(\frac{8}{10}\) chance that the third is not a spouse of the first two, and for the fourth person, there is a \(\frac{6}{9}\) probability that the fourth is not married to any of the first three.

\((\frac{10}{11})*(\frac{8}{10})*(\frac{6}{9})=\frac{16}{33}\)

Alternative explanation:

\(\frac{2*2*2*2*C_6^4}{C_{12}^4}=\frac{16}{33}\)

\(C_6^4\) - represents four couples from the given 6.

Basically, we have to take 4 couples out of 6, which can be done in \(C_6^4\) ways, and then from each couple you take one person, which can be done in 2 ways for each couple. With 4 couples, this is \(2^4=2*2*2*2\) ways.

One more approach:

Say we have couples like this (A-man, B-woman):

A1 and B1

A2 and B2

A3 and B3

A4 and B4

A5 and B5

A6 and B6

So we have 6 men (6 x A) and 6 women (6 x B)

You can pick 4 people in the following ways:
  1. 4 men - i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
  2. 3 men and 1 woman - i.e. \(C_6^3 * C_3^1\) (1 of 3 and not of 6 b/c only 3 women are left to avoid 'coupling' chosen men) = \(\frac{6!}{3!*3!} * \frac{3!}{1!*2!} = 60\)
  3. 2 men and 2 women - i.e. \(C_6^2 * C_4^2\) (again 4 women left to avoid 'coupling') = \(\frac{6!}{2!*4!} * \frac{4!}{2!*2!} = 90\)
  4. 1 man and 3 women - i.e. \(C_6^1 * C_5^3 = \frac{6!}{1!*5!} * \frac{5!}{2!*3!} = 60\)
  5. 4 women - i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)

So there are \(15+60+90+60+15=240\) possibilities.

Divide this by general possibilities of picking 4 people out of 12 - i.e. \(C_{12}^4 = \frac{12!}{4!*8!} = 495\).

\(\frac{240}{495} = \frac{16}{33}\).

Yet another approach:

Each couple can send only one "representative" to the committee. We can choose 4 couples (as there should be 4 members) to send that one "representatives" to the committee in \(C^4_6\) # of ways.

But these 4 chosen couples can send two persons (either husband or wife): \(2*2*2*2=2^4\).

So # of ways to choose 4 people out 6 married couples so that none of them would be married to each other is: \(C^4_6*2^4\).

Total # of ways to choose 4 people out of 12 is \(C^4_{12}\).

\(P=\frac{C^4_6*2^4}{C^4_{12}}=\frac{16}{33}\)


Answer: D


Hi,

Can you please tell me what's wrong with this approach?

P(none are married) = 1 - P(at least one married)
P(at least one married) = 6C1(Choose one couple out of 6)*10C2(Remaining two can be anyone) / 12C4
= (6C1* 10C2)/12C4
= 6/11

Therefore, P(none are married) = 1 - 6/11 = 5/11

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Re: M06-10 [#permalink]

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New post 06 Jul 2017, 10:10
Total ways of Selecting 4 people out of 12 = 12C4

One person out of couple and 1 from rest 5 couples (or 10 people) = 2C1 * 10C3
No of ways of selecting the couple (whether we are selecting 1st one or 2nd one etc.) = 6C1

Solution: (2C1 * 10C3 * 6C1) / 12C4 = 16/33

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Re: M06-10 [#permalink]

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New post 15 Aug 2017, 01:20
can anyone please tell why is this approach wrong ?

probability that no one is married : (12C1 * 10C1*8C1*6C1)/12C4
no of ways to select 1st person= 12C1
no of ways selecting 2nd person: 10C1 -- excluding the person selected in the 1 and his/her spouse
henceforth 8C1 and 6C1..
selecting 4 people will be 12C4.


the ans is 64/11 which is clearly against the probability rule. but i want to know what is wrong with this approach.

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M06-10 [#permalink]

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New post 25 Aug 2017, 11:25
Alternatively:

P(none) = 1 - P(one couple is selected OR two couples are selected)
(A) Total ways to select 4 from 12 = 12C4 = 495
(B) Total ways to select 1 couple from 6 = 6C1 x (10C2 - 5C1) = 6*40 = 240
i.e. imagine a couple is chosen thus {AB,_,_}: the 3rd and 4th slots can accommodate any 2 from 10 remaining members less the 5 ways (5C1) you can select a married pair from those very 10
(C) Total ways to select 2 couples from 6 = 6C2 = 15

So P(none) = 1 - (240+15)/495 = 16/33

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M06-10   [#permalink] 25 Aug 2017, 11:25
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