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?
(c) Are the events in part (a)(i) complementary or mutually exclusive or both
Answers
Using the probability concept, it is found that:
a) i) There is a 0.059 = 5.9% probability that the first person selected is either in maintenance or a secretary.
ii) There is a 0.849 = 84.9% probability that the first person selected is not in management.
b) The events are neither complementary nor mutually exclusive.
A probability is the number of desired outcomes divided by the number of total outcomes.
Item a:
In this problem, there is a total of 120 + 50 + 1460 + 302 + 68 = 2000 people.
Of those, 50 + 68 = 118 are either in maintenance or a secretary, hence:[tex]p = \frac{118}{2000} = 0.059[/tex]
There is a 0.059 = 5.9% probability that the first person selected is either in maintenance or a secretary.
Item ii:
2000 - 302 = 1698 people are not in management, hence:
[tex]p = \frac{1698}{2000} = 0.849[/tex]
There is a 0.849 = 84.9% probability that the first person selected is not in management.
Item b:
If a person is either in maintenance or a secretary, they are also not in management, hence the events are neither complementary nor mutually exclusive.
To learn more about the probability concept, you can take a look at https://brainly.com/question/24437717
Following are the solution to the given points:
For point a:
i)
[tex]\to P(B \cup E) = P(B) + P(E) - P( B\cap E)[/tex]
Where [tex]P(B \cap E) = 0[/tex] because they are both separate occasions It is impossible to be a member of both departments.
[tex]\to P(B\cup E) =\frac{50}{120 + 50 + 1460 + 302 + 68} +\frac{68}{120 + 50 + 1460+ 302 +68}\\\\[/tex]
[tex]=\frac{50}{2000} +\frac{68}{2000}\\\\=\frac{50+68}{2000} \\\\=\frac{118}{2000} \\\\=\frac{ 59}{1000} \\\\=0.059\\[/tex]
ii)
[tex]\to P(D^c) = 1 - P(D)[/tex]
[tex]=1- (\frac{302}{120+50+1460+302+68})\\\\=1- (\frac{302}{2000})\\\\=1- (\frac{ 151}{1000})\\\\= 1- 0.151 \\\\= 0.849\\\\[/tex]
For point b:
Please find the attached file.
For point c:
Events in a. I are mutually incompatible since [tex]P(B \cap E) = 0[/tex]. One would be a part of the both departments.
Learn more:
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Step-by-step explanation:
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Step-by-step explanation: