A city council consists of 9 members. Four are Republicans, three are Democrats, and two are Independents. If a committee of three is to be selected, find the probability of selecting
(d)Two Democrats and one Independent. Round your answer to five decimal places. The probability of selecting two Democrats and one Independent is

Step 1 ：The city council consists of 9 members: 4 Republicans, 3 Democrats, and 2 Independents. We are to select a committee of 3 members.

Step 2 ：The total number of ways to select a committee of three from nine members is given by the combination formula \(C(n, r) = \frac{n!}{r!(n-r)!}\), where n is the total number of items, r is the number of items to choose, and ! denotes factorial. In this case, n = 9 and r = 3. So, the total number of ways to select a committee of three is \(C(9, 3) = 84\).

Step 3 ：The number of ways to select two Democrats and one Independent is given by the product of the number of ways to select two Democrats from three and the number of ways to select one Independent from two. This is given by the formula \(C(3, 2) * C(2, 1) = 6\).

Step 4 ：The probability of selecting two Democrats and one Independent is then given by the ratio of the number of ways to select two Democrats and one Independent to the total number of ways to select a committee of three. So, the probability is \(\frac{6}{84} = 0.07142857142857142\).

Step 5 ：Rounding the final answer to five decimal places, we get \(\boxed{0.07143}\).