If $4$ people are chosen out of $6$ married couples, what is the chance that exactly one married couple is among the $4$ people?

Since there seems to be some confusion among the other answers, I'll expand my comment into an answer.

You can specify each such group of four people by choosing which three of the six couples are to be represented, then which one of the three couples is to appear in its entirety, then which one of the two people from the second couple is included, and finally which one of the two people from the third couple is included.

This makes for $${6\choose 3}\cdot{3\choose 1}\cdot{2\choose 1}\cdot{2\choose 1}=20\cdot3\cdot2\cdot2$$ favorable outcomes. Dividing by the ${12\choose4}=33\cdot15$ total outcomes gives $16/33$.

Answer:$$\frac{6.\binom52.2^2}{\binom{12}4}=\frac{16}{33}$$

Explanation of numerator:

• $6$ possibilities to choose a couple that provides $2$ selected persons.
• $\binom52$ possibilities to choose $2$ couples that provide exactly $1$ selected person.
• For the $2$ couples that provide exactly $1$ selected person there are $2$ possibilities.

Answer by two approaches

Using combinations

$\dfrac{\binom61\cdot\binom52\cdot2^2}{\binom{12}{4}}= \dfrac{16}{33} $

[ The $2^2$ is to account that either of the couple may be chosen ]

Using permutations

First pair can be chose and lined up in $6\cdot4\cdot3$ ways, and the rest of the numerator takes care that no more pair is selected.

$Pr = \dfrac{72\cdot10\cdot8}{12\cdot11\cdot10\cdot9} = \dfrac{16}{33}$