Question

A club consists of 21 Republicans and 25 Democrats. If four people are chosen at random to be on a​ committee, what is the probability that at least one person from each political party will be​ chosen?

Answer 1

Answer:

88.58% probability that at least one person from each political party will be​ chosen

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Also, the order which the people are chosen is not important. So the combinations formula is used to solve this problem.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

If four people are chosen at random to be on a​ committee, what is the probability that at least one person from each political party will be​ chosen?

Either only Republicans or Democrats are chosen, or at least one member from each party is chosen. The sum of these probabilities is 100%.

The easier way to solve this is first finding the probability that members are only from one party.

Probability of having members from only one party:

Desired outcomes:

Only Democrats:

4 from a set of 25.

Only Republican:

4 from a set of 21.

So

$D = C_{25,4} + C_{21,4} = \frac{25!}{4!21!} + \frac{21!}{4!17!} = 18635$

Total outcomes:

4 from 21+25 = 46 total members

$T = C_{46,4} = \frac{46!}{4!42!} = 163185$

Probability:

$p = \frac{D}{T} = \frac{18635}{163185} = 0.1142$

11.42% probability of having people from only one party in the committee

Probability of at least one person from each party:

p + 11.42 = 100

p = 88.58

88.58% probability that at least one person from each political party will be​ chosen