Answer:
Step-by-step explanation:
Given that a small manufacturing firm has 250 employees. Fifty have been employed for less than 5 years and 125 have been with the company for over 10 years. So remaining 75 are between 5 and 10 years.
Suppose that one employee is selected at random from a list of the employees
A) Probability that the selected employee has been with the firm less than 5 years = [tex]\frac{50}{250 } \\= 0.20[/tex]
B) Probability that the selected employee has been with the firm between 5 and 10 years
= [tex]\frac{75}{250 } \\= 0.30[/tex]
C) Probability that the selected employee has been with the firm more than 10 years
= [tex]\frac{125}{250} =0.50[/tex]
a) P(A) = 0.2
P(C) = 0.5
P(A or B) = 0.2+0.3 = 0.5
P(A and C) = 0 (since A and C are disjoint)