3+7(0.7+1.3) What is the answer and how to solve it?
1 answer:
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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 =
B) Probability that the selected employee has been with the firm between 5 and 10 years
=
C) Probability that the selected employee has been with the firm more than 10 years
=
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)