I was asked to show that $$Pr(A|B) = 1 - Pr(A|B^c)$$ but I cannot find a relation between them, I also saw that $$Pr(A|B) = 1 - Pr(A^c|B)$$ so I'm not really sure about the equation that i was asked to proof, if anyone has any advice would be great.
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$\begingroup$The first one is false.
The second expression is true.
To prove: $P(A|B) = 1-P(A^c|B)$
By definition:
$P(A|B) = \frac{P(AB)}{P(B)}$
$1 - P(A^c|B) = 1 - \frac{P(A^cB)}{P(B)}$
So you have to prove:
$\frac{P(AB)}{P(B)} = 1 - \frac{P(A^cB)}{P(B)}$
Multiply through by $P(B)$ and get
$P(AB) = P(B) - P(A^cB)$
$P(AB) + P(A^cB) = P(B)$
Since the two terms on the LHS are mutually exclusive you can do:
$P(AB \cup A^cB) = P(B)$
And that completes the proof.
$\endgroup$ $\begingroup$Here's an easy way to show that the top equation is false: consider the case where $A$ and $B$ are independent, whence $\mathbb P(A \mid B) = \mathbb P(A \mid B^c) = \mathbb P(A)$. If $\mathbb P(A) \neq 1/2$, you have a problem.
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