A city council consists of five democrats and six republicans. If a committee of six people is selected, find the probability of selecting two democrats and four republicans.
We can use the combination formula to solve this problem.
The total number of people on the city council is 5 + 6 = 11.
We want to select a committee of 6 people.
The number of ways to select 2 democrats from 5 is denoted by C(5, 2), which is calculated as follows:
C(5, 2) = 5! / ((5-2)! * 2!) = 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10.
Similarly, the number of ways to select 4 republicans from 6 is calculated as follows:
C(6, 4) = 6! / ((6-4)! * 4!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15.
Therefore, the total number of ways to select the committee with two democrats and four republicans is C(5, 2) * C(6, 4) = 10 * 15 = 150.
The total number of ways to select a committee of 6 people from a group of 11 is C(11, 6), which is calculated as follows:
C(11, 6) = 11! / ((11-6)! * 6!) = 11! / (5! * 6!) = (11 * 10 * 9 * 8 * 7) / (5 * 4 * 3 * 2 * 1) = 462.
So, the probability of selecting two democrats and four republicans is given by:
P(2democrats 4republicans) = (number of ways to select 2 democrats and 4 republicans) / (total number of ways to select a committee of 6 people)
P(2democrats 4republicans) = 150 / 462 ≈ 0.3247.
Therefore, the probability of selecting two democrats and four republicans is approximately 0.3247 or 32.47%.