Question

a group consists of four democrats and five republicans. four people are selected to attend a conference. a. in how many ways can four people be selected from this group of nine? b. in how many ways can four republicans be selected from the five republicans? c. find the probability that the selected group will consist of all republicans.

Explanation:

Step1: Recall combination formula

The combination formula is $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n$ is the total number of items and $r$ is the number of items to be chosen.

Step2: Calculate number of ways to select 4 people from 9 (for part a)

Here $n = 9$ and $r=4$. So $C(9,4)=\frac{9!}{4!(9 - 4)!}=\frac{9!}{4!5!}=\frac{9\times8\times7\times6}{4\times3\times2\times1}=126$.

Step3: Calculate number of ways to select 4 Republicans from 5 (for part b)

Here $n = 5$ and $r = 4$. So $C(5,4)=\frac{5!}{4!(5 - 4)!}=\frac{5!}{4!1!}=5$.

Step4: Calculate probability that selected group consists of all - Republicans (for part c)

The total number of ways to select 4 people from 9 is $C(9,4)=126$ (from part a), and the number of ways to select 4 Republicans from 5 is $C(5,4)=5$ (from part b). The probability $P=\frac{C(5,4)}{C(9,4)}=\frac{5}{126}$.

Answer:

a. 126
b. 5
c. $\frac{5}{126}$