Question

Choose an employee person at random. Let A be the event that the person is a female and B be the event that the person holds a managerial position. Data from the US department of labor suggests that P(A)= 0.47 and P(B|A)= 0.34

A) explain what P(A)= 0.47 means in context
B) explain what P(B|A)= 0.34 means in context
C) what is the probability that a randomly chosen employed person is a male?
D) what is the probability that a randomly chosen employed person is a female manager?
E) what is the probability that a randomly chosen employed female is not a manager?

Answer 1

Hello there.

E) what is the probability that a randomly chosen employed female is not a manager?

0.3102.

Answer 2

A: Female B: Managerial Position

P(A) = 0.47

P(B|A) = 0.34

Find P(A') -> [Probability that the person is male]

By the complement rule

P(A') = 1 - P(A)

= 1 - 0.47

= 0.53

P(A') = 0.53

So the probability of a male being chosen is 0.53.

2.) P(B|A) = P( B and A) / P(A)

Use the values given P(A) and P(B|A) to find P(B and A) which will be your answer

3.) What is the probability that a randomly chosen employed female is not a manager ?

What is the probability that a female is not a manager

Find P(A and B')

Step #1, First Find P(B' | A).

It can be shown that

P(B | A ) + P(B' | A) = 1 [ You can verify this with a Venn diagram]

So

0.34 + P(B' | A) = 1

so

P(B' | A) = 0.66

P(B' | A) = P(B' and A) / P(A)

0.66 = P(B' and A) / 0.47

So P(B' and A) = 0.3102

So the chance you will be someone that is female and not a manager is 0.3102.