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?

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.

Hello there.

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

0.3102.

C. because 6÷3=2 so 3 is half of six so 3/6