Question

Consider a political discussion group consist of five Democrats and six Republicans. Four people are selected (at random) to attend a conference. Find the probability that the selected group will consist of all Republicans? Round to the nearest thousandth

Answer 1

In how many ways can four people be selected from this group of eleven?

$$\dbinom{11}{4}={11! \over 4!(11-4)!}$$

 In how many ways can four Republicans be selected from the six Republicans?

$$\dbinom{6}{4}={6! \over 4!(6-4)!}$$

In how many ways can zero Democrats be selected from the five Democrats?

$$\dbinom{5}{0}=1$$

Find the probability that the selected group will consist of all Republicans

$$P={{\dbinom{6}{4}\dbinom{5}{0}} \over {\dbinom{11}{4}}}={{6! (1)4!(11-4)!}\over{ 4!(6-4)!11!}}={1 \over 22}=0.045$$