A city council consists of seven democrats and six republicans. If a committee of six people is selected, find the probability of selecting four Democrats and two Republicans.

Answer 1

To find the probability of selecting four Democrats and two Republicans, we can use the concept of combinations.

The total number of ways to select a committee of six people from thirteen (7 Democrats + 6 Republicans) is given by the combination formula:

C(13, 6) = 13! / (6! * (13-6)!) = 13! / (6! * 7!) = (13 * 12 * 11 * 10 * 9 * 8) / (6 * 5 * 4 * 3 * 2 * 1) = 1716

The number of ways to select four Democrats from seven is given by the combination:

C(7, 4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35

The number of ways to select two Republicans from six is given by the combination:

C(6, 2) = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15

Therefore, the number of ways to select four Democrats and two Republicans is given by:

C(7, 4) * C(6, 2) = 35 * 15 = 525

The probability of selecting four Democrats and two Republicans is:

P(Four Democrats and Two Republicans) = 525 / 1716 ≈ 0.3056

So, the probability of selecting four Democrats and two Republicans is approximately 0.3056 or 30.56%.