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%.
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.
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