Question

Given that the P(A) = 0.32, P(B) = 0.34, and P(AUB)= 0.56, find the P( B ' | A' ).

Answer 1

p(A' intersection B')=p(A)+p(B)-p(A intersection B)
p(A intersection B) u can get from the equations

Answer 2

P( B ' | A' ).=P(A' intersection B')/P(B')

Answer 3

can you solve it now

Answer 4

\[P(A'\cap B')=1-P(A\cup B)\]

\[P(B'|A')=\frac{P(B'\cap A')}{P(A')}\]

Answer 5

thus

\[P(B'|A')=\frac{P(B'\cap A')}{P(A')}=\frac{1-P(A\cup B)}{1-P(A)}\]