What is the probability that no married couples are among the chosen? Eight married couples are standing in a room.
4 people are randomly chosen.
What is the probability that no married couples are among the chosen?

Start by choosing a person. Then choose a second. The probability that it is not a partner of the first is $\frac{14}{15}$. Suppose that this is indeed the case and choose a third. In that case the probability that it is not a partner of one of the two that are allready chosen is $\frac{12}{14}$. Suppose again that this is indeed the case and finally choose the fourth. In that case the probability that it is not a partner of one of the three that are allready chosen is $\frac{10}{13}$.
This leads to a probability of $$\frac{14}{15}\times \frac{12}{14}\times \frac{10}{13}=\frac{8}{13}$$

The same principle in a more formal jacket:
Choose persons one by one. 
For $n\in\left\{ 1,2,3,4\right\} $ define
$X_{n}$ as the number of married couples under the first $n$ persons
that are chosen. 
If $n$ persons have been chosen in such a way that $X_n=0$ then $16-n$ are left and among them there are exactly $16-2n$ that are not partner of one that is
allready chosen. 
So: $$P\left\{ X_{n+1}=0\mid X_{n}=0\right\} =\frac{16-2n}{16-n}$$ and, since $X_{n+1}=0\Rightarrow X_n=0$, consequently: $$P\left\{ X_{n+1}=0\right\} =\frac{16-2n}{16-n}P\left\{ X_{n}=0\right\} $$ 
leading to:
$$P\left\{ X_{4}=0\right\} =\frac{10}{13}P\left\{ X_{3}=0\right\} =\frac{10}{13}\frac{12}{14}P\left\{ X_{2}=0\right\} =\frac{10}{13}\frac{12}{14}\frac{14}{15}P\left\{ X_{1}=0\right\} =\frac{10}{13}\frac{12}{14}\frac{14}{15}=\frac{8}{13}$$
