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If we have the probabilities of $P(A)$, $P(B)$ and $P(A\mid B)$, how can we calculate the probability of $P(A\mid\neg B)$ ?

Does $A$ depends on $\neg B$ if it Actually depends on $B$ ?

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2 Answers 2

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Well, not $B$ is just $B^C$ so from the formula,

$$P(A\mid B^C) = \frac{P(A \cap B^C)}{P(B^C)}$$

You can calculate $P(B^C) = 1 - P(B)$ and we can get $P(A \cap B) = P(A\mid B)P(B)$ which can give us $P(A \cap B^C)$ using

$$P(A) = P(A \cap B) \cup P(A \cap B^C)$$

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  • $\begingroup$ I didn't understand the last part. $\endgroup$
    – reaffer
    Dec 27, 2015 at 22:27
  • $\begingroup$ Added it into the post. $\endgroup$
    – Future
    Dec 27, 2015 at 22:31
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Take a glance on the Venn Diagram:

enter image description here

Let $a=\mathbb P(A), b=\mathbb P(B), r=\mathbb P(A\cap B)$.

Hence, $\mathbb P(A|B)=\frac rb$ and also $\mathbb P(A|B^c)=\frac{a-r}{1-b}$.

So, given $a,b,\frac rb=x$

$$P(A|B^c)=\frac{a-r}{1-b}=\frac{a-bx}{1-b}$$

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