Probability of A given not B

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So I have been given the following details:

$p(a\mid b)=0.2\\ p(a\mid \lnot b)= 0.6$

How would I use this to determine $p(\lnot a\mid b)$? Is there some kind of a formula/procedure I can use or is there incomplete information?

I don't think that $p(\lnot a\mid b)= 1-p(a\mid b)$, but I can't think of any other ways

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Once you draw the probability tree and let P(b)=x, it will become clear to you.

Given b, either a or (not a) will happen for sure. Thus, $P(a|b)+P(not\; a|b)=1$ for sure.