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 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 0.30000.
part 4 of 5
(a) two democrats and one independent. round your answer to five decimal places.
the probability of selecting two democrats and one independent is

Explanation:

Step1: Calculate the number of ways to choose 2 Democrats, 1 Independent

The number of ways to choose \(r\) items from \(n\) items is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\).
The number of ways to choose 2 Democrats out of 3 is \(C(3,2)=\frac{3!}{2!(3 - 2)!}=\frac{3!}{2!1!}=\frac{3\times2!}{2!×1}=3\).
The number of ways to choose 1 Independent out of 3 is \(C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=\frac{3\times2!}{2!×1}=3\).
The total number of ways to choose 3 members out of 10 is \(C(10,3)=\frac{10!}{3!(10 - 3)!}=\frac{10\times9\times8}{3\times2\times1}=120\).

Step2: Calculate the probability

The probability \(P\) is the number of favorable outcomes divided by the number of total outcomes.
The number of favorable outcomes (choosing 2 Democrats and 1 Independent) is \(C(3,2)\times C(3,1)=3\times3 = 9\).
So the probability \(P=\frac{9}{120}=0.07500\).

Answer:

\(0.07500\)