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Events A and B are such that P(A)=0.7, P(B)=0.2, and P(A∩B)=0.2. Find P(A|B').
I found out that P(A u B) = 0.7, but I'm not sure how to work out the conditional probability - I've tried using the formula and I got P(A|B') = (0.7*0.8)/0.8, but that seems wrong.
2 Answers
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$$P(A\mid B')=\frac{P(A\cap B')}{P(B')}=\frac{P(A)-P(A\cap B)}{1-P(B)}$$
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Hint:
$$P(A) = P(A\cap \Omega) = P(A\cap (B\cup B')) = P((A\cap B)\cup (A\cap B')) = P(A\cap B) + P(A\cap B')$$
$\Omega$ here represents the probability space.
We know that $P(A)=0.7$ and we know that $P(A\cap B)=0.2$ so...
