Question 1190835
<font color=black size=3>
You only really need to memorize or write down on a reference sheet one formula, and it is
P(A|B) = [P(B|A)*P(A)]/[P(B|A)*P(A)+P(B|¬A)*P(¬A)]


This is because the second formula is derived from the first one. 


Notice how replacing every A with ¬A gets us this:
P(¬A|B) = [P(B|¬A)*P(¬A)]/[P(B|¬A)*P(¬A)+P(B|¬¬A)*P(¬¬A)]
If you see ¬¬A, then that's the same as A
P(¬A|B) = [P(B|¬A)*P(¬A)]/[P(B|¬A)*P(¬A)+P(B|A)*P(A)]


Now if you replaced every B with ¬B, then you should end up with the second formula you mentioned.


Use this idea to compute what P(A|¬B) would be. 
You would of course need to start back over with the P(A|B) formula. Then replace every B with ¬B and simplify any ¬¬B into B.


Side note:
P(B) = P(B|A)*P(A) + P(B|¬A)*P(¬A)
due to the law of total probability
</font>