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?
 A: To get the probability, you'll need


*

*the total number of combinations of $16$ people taken $4$ at a time, and

*the total number of those combinations that have no married couples among them.


You have the first part:  ${16 \choose 4}$.
Rather than count up the number of combinations for the second part directly, you can count the complement, and subtract from the total.
So if you do have at least one married couple in the chosen, you can have one couple, or two.
If you have two married couples in the chosen group, then there are ${8 \choose 2}$ possible combinations.
If you have (exactly) one married couple, first you pick the married couple in the group: ${8 \choose 1}$.  Now, from the remaining $14$ people, you pick two that aren't married to each other.  So pick the two couples that you're going to draw from -- ${7 \choose 2}$ -- and then pick the particular spouse from each one -- ${2 \choose 1} \cdot {2 \choose 1}$.
Now, you subtract the number of combinations that do have at least one married couple from the total number of combinations to get the number of combinations that have no married couples.  This gives as your probability:
$$P = \frac{{16 \choose 4} - {8 \choose 2} - {8 \choose 1}{7 \choose 2}{2 \choose 1}{2 \choose 1}}{{16 \choose 4}} = \frac{8}{13} \approx 0.61538.$$
A: Another way to figure this out would be to choose 4 different married couples to provide a chosen person (there are $\left(\begin{array}{c} 8\\ 4\end{array}\right)=70$ ways to do this).  Then pick one member of each of the 4 chosen couples (there are $2^4=16$ ways to do this).  So altogther there are $70\cdot 16=1120$ ways to choose 4 people so that no two are married to each other.  
So the probability of none married to each other among the 4 chosen is
$\frac{1120}{\left(\begin{array}{c} 16\\ 4\end{array}\right)}=\frac{1120}{1820}=\frac8{13}$.  
So same as John's answer, just a different approach to counting in the problem.
A: 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:
$$P\left\{ X_{4}=0\right\} =\frac{10}{13}P\left\{ X_{3}=0\right\} =\frac{10}{13}\frac{12}{14}P\left\{ X_{2}=0\right\} =\frac{10}{13}\frac{12}{14}\frac{14}{15}P\left\{ X_{1}=0\right\} =\frac{10}{13}\frac{12}{14}\frac{14}{15}=\frac{8}{13}$$
A: For no married couples: choose 4 from the 8 available couples, considering that within each couple, we can have either one of the two partners.
$\binom{8}{4} \times 2^4$
Now divide this by the total number of selections we can make from 16 people:
$$\frac{\binom{8}{4} \times 2^4}{\binom{16}{4}} = \frac{8}{13} \approx 0.6154 $$
A: Here's yet one more way of getting the answer.  The number of ways of making a selection with $m$ men and $4-m$ women that avoid married couples is ${8\choose m}{8-m\choose4-m}$.  The requested probability is therefore
$${{8\choose4}{8\choose0}+{8\choose3}{5\choose1}+{8\choose2}{6\choose2}+{8\choose1}{7\choose3}+{8\choose0}{8\choose4}\over{16\choose4}}={1120\over1820}={8\over13}$$
A: Very easy way to do this is to think of it as the following:
you have 8 couples, which means 16 members,
since you have to pick 4 people that are not married
you have the following possibilities: 16 * 14 * 12 * 10
this is because when you select the first person, you can't take its lover.
now you can compute the probability easily as 16 * 14 * 12 * 10 / 16 * 15 * 14 *13 = 8/13
