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:

- 4 men - i.e. \(C_6^4 = \frac{6!}{2!*4!} = 15\)
- 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) = \(\frac{6!}{3!*3!} * \frac{3!}{1!*2!} = 60\)
- 2 men and 2 women - i.e. \(C_6^2 * C_4^2\) (again 4 women left to avoid 'coupling') = \(\frac{6!}{2!*4!} * \frac{4!}{2!*2!} = 90\)
- 1 man and 3 women - i.e. \(C_6^1 * C_5^3 = \frac{6!}{1!*5!} * \frac{5!}{2!*3!} = 60\)
- 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

P(at least one married) = 6C1(Choose one couple out of 6)*10C2(Remaining two can be anyone) / 12C4

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