A city council consists of seven democrats and seven republicans. If a committee of five people is selected, find the probability of selecting two democrats and three republicans.
The total number of ways to select a committee of five people from a group of fourteen (seven Democrats and seven Republicans) is given by the combination formula:
C(14, 5) = 14! / (5!(14-5)!) = (14*13*12*11*10)/(5*4*3*2*1) = 2002.
Now we need to determine the number of ways to select two Democrats and three Republicans. This can be calculated by multiplying the number of ways to choose two Democrats from seven (C(7, 2) = 7! / (2!(7-2)!)) by the number of ways to choose three Republicans from seven (C(7, 3) = 7! / (3!(7-3)!)).
C(7, 2) * C(7, 3) = (7! / (2!(7-2)!)) * (7! / (3!(7-3)!))
= (7*6*5!) / (2*5!) * (7*6*5!) / (3*4*3!)
= (7*6) / (2*1) * (7*6*5*4) / (3*2*1)
= 21 * 35
= 735.
So, the number of ways to select two Democrats and three Republicans is 735.
Finally, the probability of selecting two Democrats and three Republicans is given by:
P(2D, 3R) = Number of ways to select two Democrats and three Republicans / Total number of ways to select a committee of five people
= 735 / 2002
≈ 0.366.