Question

a city council consists of 10 members. four are republicans, three are democrats, and three are independents. if a committee of three is to be selected, find the probability of selecting.
part 1 of 5
(a) all republicans. round your answer to five decimal places.
the probability of selecting all republicans is 0.03333.
part 2 of 5
(b) all democrats. round your answer to five decimal places.
the probability of selecting all democrats is 0.00833.
part: 2 / 5
part 3 of 5
(c) one of each party. round your answer to five decimal places.
the probability of selecting one person from each party is

Explanation:

Step1: Calculate the number of ways to choose 1 from each party

We use the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\).
For Republicans: \(C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\)
For Democrats: \(C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\)
For Independents: \(C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\)
The number of ways to choose 1 from each party is \(C(4,1)\times C(3,1)\times C(3,1)=4\times3\times3 = 36\)

Step2: Calculate the total number of ways to choose 3 from 10

The total number of ways to choose 3 from 10 is \(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\)

Step3: Calculate the probability

The probability \(P=\frac{36}{120}=0.3\)

Answer:

\(0.30000\)