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A sample of employees of Worldwide Enterprises is to be surveyed about a new health care plan. The employees are classified as follows: Classification Event Number of Employees Supervisors A 120 Maintenance B 50 Production C 1,460 Management D 302 Secretarial E 68 (a) What is the probability that the first person selected is: (i) either in maintenance or a secretary? (ii) not in management?

Question

A sample of employees of Worldwide Enterprises is to be surveyed about a new health care plan. The employees are classified as follows:

Classification Event Number of Employees Supervisors A 120 Maintenance B 50 Production C 1,460 Management D 302 Secretarial E 68 (a) What is the probability that the first person selected is: (i) either in maintenance or a secretary? (ii) not in management?

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Solution

(i) To find the probability that the first person selected is either in maintenance or a secretary, we first need to find the total number of employees.

The total number of employees is 120 (Supervisors) + 50 (Maintenance) + 1,460 (Production) + 302 (Management) + 68 (Secretarial) = 2,000 employees.

The number of employees in maintenance or secretarial is 50 (Maintenance) + 68 (Secretarial) = 118 employees.

So, the probability that the first person selected is either in maintenance or a secretary is 118/2,000 = 0.059 or 5.9%.

(ii) To find the probability that the first person selected is not in management, we subtract the number of management employees from the total number of employees.

The number of employees not in management is 2,000 (Total) - 302 (Management) = 1,698 employees.

So, the probability that the first person selected is not in management is 1,698/2,000 = 0.849 or 84.9%.

This problem has been solved

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