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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.

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  • $\begingroup$ Do you know the definition of conditional probability? Do you know $P(B')$? $\endgroup$
    – aduh
    Commented Sep 26, 2016 at 1:35
  • $\begingroup$ P(B') = 1 - P(B) = 1 - 0.2 = 0.8 $\endgroup$
    – JC1
    Commented Sep 26, 2016 at 1:36
  • $\begingroup$ Good. You'll need the definition of conditional probability: $$P(A \mid B') = \frac{P(A \cap B')}{P(B')}$$ and you should be good to go. $\endgroup$
    – aduh
    Commented Sep 26, 2016 at 1:38
  • $\begingroup$ I'm not seeing your tex code as proper symbols. Is there a plugin/program I should be installing? $\endgroup$
    – JC1
    Commented Sep 26, 2016 at 1:40
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    $\begingroup$ Not in general. Only if $A$ and $B$ are independent. Seems like you need to review your textbook or course notes. $\endgroup$
    – aduh
    Commented Sep 26, 2016 at 1:45

2 Answers 2

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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...

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  • $\begingroup$ I'm not sure why this received a downvote. It seemed to me that the only piece of information the OP was missing to finish solving was how to find $P(A\cap B')$, which the above work shows that $P(A\cap B')=P(A)-P(A\cap B)$, just as the other answer said without proof. $\endgroup$
    – JMoravitz
    Commented Sep 26, 2016 at 2:14

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