If P(B) = 0.05, P(A|B) = 0.80, P(B′) = 0.95, and P(A|B′) = 0.40, find P(B|A). Be careful P(B') is not equal to 0.95; you can compute P(B') by yourself using the other info in the problem.

Answer 1

P(B) = .05, P(A)=.1-.5= .5

P(A/B) = .8

P(~B) = 1-P(B) = 1-.05= .95

P(A/~B)= .4

find P(~B) = 1-P(B) = .95

hard to escape that conclusion, in any scenario

unless there is a 3rd part C that's not A or B

then P(~B) = 1-P(B)-P(C) = not .95

but use the other info, and use Baye's Theorem

there are some surprising suprise probabilities after using Baye's Theorem

or draw a Venn diagram

P(A/B)P(B) = P(B/A)P(A)

.8(.05)= .04= P(B/A)/.5

P(B/A) = .04/.5 = .08

P(A/B)