How do I find P(A+B) given only P(A) and P(B|not A)

Question

I need to find the value of P(A+B) in terms of P(A) and P(B|Ac) where Ac is the complement of A, A+B is A or B, and B|Ac is B given A complement.

So far I've worked it out to P(A+B) = P(A)^2 + P(A)P(B) - P(B) + P(B|Ac) but can't get any further.

Answer 1

Let $x=P(A)$ and $y=P(B|\neg A)$.

$P(\neg A)=1-x$

$P(\neg B|\neg A)=1-P(B|\neg A)=1-y$

$P(\neg A \wedge \neg B) = P(\neg A) \cdot P(\neg B|\neg A) = (1-x)(1-y)$

$P(A+B)=1-P(\neg A \wedge \neg B)=1-(1-x)(1-y)=x+y-xy$

Answer 2

$$ \begin{eqnarray*} P(A \cup B)&=&P(A)+P(B)-P(A\cap B) \\ &=&P(A)+P(B\cap A)+P(B\cap \bar A)-P(A\cap B) \\ &=&P(A)+P(B\cap \bar A) \\ &=&P(A)+P(B|\bar A)P(\bar A) \\ &=&P(A)+P(B|\bar A)(1-P( A)) \end{eqnarray*} $$