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
(e)One Independent and two Republicans. Round your answer to five decimal places.
The probability of selecting one Independent and two Republicans is

Step 1 ：The city council consists of 9 members: 4 Republicans, 3 Democrats, and 2 Independents.

Step 2 ：We are to select a committee of 3 members.

Step 3 ：We want to find the probability of selecting one Independent and two Republicans.

Step 4 ：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 is $C(9,3)=84$.

Step 5 ：The number of ways to select one Independent and two Republicans is given by the product of the number of ways to select one Independent from two and the number of ways to select two Republicans from four. This is given by the product of the combination formulas $C(2,1)$ and $C(4,2)$. So, the number of ways to select one Independent and two Republicans is $C(2,1)\times C(4,2)=12$.

Step 6 ：The probability of selecting one Independent and two Republicans is then given by the ratio of the number of ways to select one Independent and two Republicans to the total number of ways to select a committee of three. So, the probability is $\frac{12}{84}=0.14285714285714285$.

Step 7 ：Final Answer: The probability of selecting one Independent and two Republicans is $\boxed{{\displaystyle 0.14286}}$.