[Math] the probability that no married couples are among the chosen

probability

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?

Best Answer

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:

$$P = \frac{{16 \choose 4} - {8 \choose 2} - {8 \choose 1}{7 \choose 2}{2 \choose 1}{2 \choose 1}}{{16 \choose 4}} = \frac{8}{13} \approx 0.61538.$$

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