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 0.07500.
part 4 / 5
part 5 of 5
(a) one independent and two republicans. round your answer to five decimal places.
the probability of selecting one independent and two republicans is

Explanation:

Step1: Calculate the combination for each part

For part (a), the number of ways to choose 3 Republicans out of 4 is \(C(4,3)=\frac{4!}{3!(4 - 3)!}=4\). The total number of ways to choose 3 members out of 10 is \(C(10,3)=\frac{10!}{3!(10 - 3)!}=120\). The probability \(P=\frac{C(4,3)}{C(10,3)}=\frac{4}{120}\approx0.03333\).
For part (b), the number of ways to choose 3 Democrats out of 3 is \(C(3,3) = 1\). The probability \(P=\frac{C(3,3)}{C(10,3)}=\frac{1}{120}\approx0.00833\).
For part (c), the number of ways to choose 1 from each party: \(C(4,1)\times C(3,1)\times C(3,1)=4\times3\times3 = 36\). The probability \(P=\frac{36}{120}=0.30000\).
For part (a) of part 4, the number of ways to choose 2 Democrats out of 3 is \(C(3,2)=\frac{3!}{2!(3 - 2)!}=3\), the number of ways to choose 1 Independent out of 3 is \(C(3,1)=3\). The total number of favorable combinations is \(C(3,2)\times C(3,1)=3\times3 = 9\). The probability \(P=\frac{9}{120}=0.07500\).
For part (a) of part 5, the number of ways to choose 1 Independent out of 3 is \(C(3,1) = 3\), the number of ways to choose 2 Republicans out of 4 is \(C(4,2)=\frac{4!}{2!(4 - 2)!}=6\). The total number of favorable combinations is \(C(3,1)\times C(4,2)=3\times6=18\). The probability \(P=\frac{18}{120}=0.15000\).

Answer:

(a) \(0.03333\)
(b) \(0.00833\)
(c) \(0.30000\)
(a) of part 4 \(0.07500\)
(a) of part 5 \(0.15000\)