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.
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%.