Question

six sales representatives for a company are to be chosen at random to participate in a training program. the company has eight sales representatives, two in each of four regions. what is the probability that the six sales representatives chosen to participate in the training program will be from only three of the four regions?
the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the total number of ways to choose 6 sales representatives out of 8

Use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 8\) and \(r=6\).

$$C(8,6)=\frac{8!}{6!(8 - 6)!}=\frac{8\times7}{2\times1}=28$$

Step2: Calculate the number of ways to choose 6 sales - representatives from 3 regions

First, choose 3 regions out of 4: \(C(4,3)=\frac{4!}{3!(4 - 3)!}=4\)
For each of the 3 chosen regions, we need to take 2 representatives from each region. But since we are choosing 6 representatives from 3 regions (2 per region), the number of ways to choose representatives from the 3 chosen regions is \(C(2,2)\times C(2,2)\times C(2,2) = 1\) (because \(C(2,2)=1\))

Answer:

\(\frac{4}{28}\approx0.1429\)