How can you compute P(not B/A) if you know P(A/not B) = .33?

Question

I know these statements are not equal (probability fallacy).

Answer 1

I don't believe that can be computed with the information given.

P(not B | A) = P(not B and A)/P(A)

or by Bayes' rule,

= P(A | not B)P(not B)/{P(A | B)P(B)+P(A | not B)P(not B)}

I really do think we would need more information. If we knew P(B) and P(A | B), we could get it using Bayes' rule.

For example, if P(B) = 0.4 and P(A | B) = 0.1, then

P(not B | A) = (0.33*0.6)/(0.33*0.6+0.1*0.4) = 0.832

But if we change to P(B) = 0.1, then

P(not B | A) = (0.33*0.9)/(0.33*0.9+0.1*0.1) = 0.967

So it does depend on those numbers. You can't get it without more information.

Doug, your supposition is incorrect. P(not A | B) = 1 - P(A | B), not 1-P(not B | A).

Answer 2

Since the probability of A + B is 1 (in the probability space defined by A and B), then P(not A/B) is equal to
1 - P(not B/A).


Doug