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?

Question: 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?

1. \(\frac{1}{33}\)
2. \(\frac{2}{33}\)
3. \(\frac{1}{3}\)
4. \(\frac{16}{33}\)
5. \(\frac{11}{12}\)

“ 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?” - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Quantitative Review". GMAT Quant section consists of a total of 31 questions. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Answer:

As we have 2 couples, so this can be arranged in the following manner: (X- Man, Y- Woman)

\(X ^1\)and \(Y^1\)

\(X^2\) and \(Y^2\)

\(X ^3\)and \( Y^3\)

\(X ^4\)and \(Y^4\)

\(X^5\) and \(Y^5\)

\(X ^6\)and \(Y^6\)

So, we can conclude that we have 6 men and 6 women (Total 12 people)

Now, we will write down the number of ways in which 4 people can be selected:

1. Selection of all 4 men =\(^6C_4 \) =\(\frac{6!}{4!(6-4)!} \) = 15 ways
2. Selection of 3 men and 1 woman =\(^6C_3*^3C_1 \) = \(\frac{6!}{3!(6-3)!}\frac{3!}{1!(3-1)!} \)= 60 ways
3. Selection of 2 men and 2 women = \(^6C_2*^4C_2 \)=\(\frac{6!}{2!(6-2)!}\frac{4!}{2!(4-2)!} \) = 90 ways
4. Selection of 1 men ad 3 women =\(^6C_1*^5C_3 \) =\(\frac{6!}{1!(6-1)!}\frac{5!}{3!(5-3)!} \) = 60 ways
5. Selection of 4 women = \(^6C_4 \)=\(\frac{6!}{4!(6-4)!} \) = 15 ways

Hence, Total number of possibilities = 15 + 60 + 90 + 60 + 15 = 240 possibilities.

Now, we will calculate the general possibilities of picking out 4 people from the 12 people

General possibilities =\(^{12}C_4 \) =\(\frac{12!}{4!(12-4)!} \) = 495

Therefore, Required probability =\(\frac{Total In mber of Possibilites}{General Possibilites}\) =\(\frac{240}{495}\) =\(\frac{16}{33}\)

Correct Answer: D

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