A city council consists of eight democrats and eight republicans. If a committee of five people is selected, find the probability of selecting two democrats and three republicans.
First, we need to find the total number of ways to select a committee of five people from a city council of 16 members. This can be done using combinations. The formula for combinations is:
nCr = n! / (r!(n-r)!)
where n is the total number of items and r is the number of items to be selected.
In this case, n = 16 (eight democrats + eight republicans) and r = 5 (size of the committee).
So, the total number of ways to select a committee of five from the city council is:
16C5 = 16! / (5!(16-5)!) = (16 * 15 * 14 * 13 * 12) / (5 * 4 * 3 * 2 * 1) = 4368.
Next, we need to find the number of ways to select two democrats and three republicans. There are 8 democrats and we need to choose 2 of them, which can be done in 8C2 = 8! / (2!(8-2)!) = (8 * 7) / (2 * 1) = 28 ways. Similarly, there are 8 republicans and we need to choose 3 of them, which can be done in 8C3 = 8! / (3!(8-3)!) = (8 * 7 * 6) / (3 * 2 * 1) = 56 ways.
So, the number of ways to select two democrats and three republicans is 28 * 56 = 1568.
Finally, the probability of selecting two democrats and three republicans is the number of favorable outcomes (1568) divided by the total number of possible outcomes (4368):
P(2 democrats, 3 republicans) = 1568 / 4368 ≈ 0.359