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 = $\frac{50}{250 } \\= 0.20$

B) Probability that the selected employee has been with the firm between 5 and 10 years

= $\frac{75}{250 } \\= 0.30$

C) Probability that the selected employee has been with the firm more than 10 years

= $\frac{125}{250} =0.50$

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)

7 0

3 years ago

PLEASE HELP ME!!!!!!!!!!!

dusya [7]

Answer:

he was being cheated

Step-by-step explanation:

3 0

3 years ago