If ​P(B)=0.4​, P(A|B)=0.5, P(B′​)=0.6, and​ P(A|B′)=0.7, what is P(B|A)?

Question

Statistics probability problem.

I could really use some help figuring out how to solve this.

Answer 1

P(A)= P(A∩S)

      = P[A∩(BUB')]

      = P[(A∩B) U (A∩B')]

      = P(A∩B) + P(A∩B')              {(A∩B) and (A∩B') are mutually exclusive}

      = P(A|B)P(B) + P(A|B')P(B')

      = (0.5)(0.4) + (0.7)(0.6)

      = 0.62

 

P(B|A) = P(A∩B)/P(A)

           = 0.2/0.62

           = 0.32

Answer 2

I conclude that P(B) = 0.4 based on P(B') = 0.6

 

From P(A|B') = 0.7 I write 0.7 = P(A∩B') / P(B') and then conclude P(A∩B')  = 0.7(0.6) = 0.42

 

If I label two semi-overlapping circles A and B, and use 

a = part of A not in intersection with B

b = part of B not in intersection with A

c =  A ∩ B

d = part outside both circles 

then A∩B' = {a,c} ∩ {a,d} = {a) and this leads to a contradiction because a cannot equal both 0.4 as well as 0.42

 

I WILL BE GLAD TO SEE ANALYSIS OF THIS FROM ANY INTERESTED TUTOR.