Question

1. Of 6 employees, 3 have been with the company five or more years. If 4 employees are chosen randomly from the group of 6, what is the probability that exactly 2 will have five or more years seniority

Answer 1

Answer:

Probability = 0.6

Step-by-step explanation:

We are given that out of 6 employees, 3 have been with the company for five or more years which means remaining 3 employees have been with the company for less than 5 years.

We have to find the probability that If 4 employees are chosen randomly from the group of 6, what is the probability that exactly 2 will have five or more years seniority;

This means that 2 employees will be selected from the group of 3 employees who are with the company for five or more years and remaining 2 employees will be selected from the group of 3 employees who have been with the company for less than 5 years.

Number of ways in which exactly 2 will have five or more years seniority

= $^{3} C_2 * ^{3} C_2$

Total number of ways for selecting 4 employees from group of 6 = $^{6} C_4$

So, required probability = $\frac{^{3} C_2*^{3} C_2}{^{6} C_4}$ = $\frac{3!}{2!*1!}* \frac{3!}{2!*1!} * \frac{4!*2!}{6!}$   {$\because ^{n}C_r = \frac{n!}{r!*(n-r)!}$ }

                                        = $\frac{3*2!}{2!}*\frac{3*2!}{2!}*\frac{4!*2!}{6*5*4!}$ = $\frac{18}{30}$ = 0.6