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The probability of $B$ given $A$ and one minus the probablity of $\sim B$ given $A$.


$$P(B|A)=1-P(\sim B|A)$$

The events are not necessarily mutually exclusive, I just wanted to know if the statements are equivalent?

I just started statistics so I don't know how much more detail I can give, please let me know in your comments.

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you mean \ or / ? – HipsterMathematician Sep 21 '12 at 0:23
@MeAndMath What are you talking about? Are you talking about Latex code? – yiyi Sep 21 '12 at 0:25
yeah,yeah, bad. – HipsterMathematician Sep 21 '12 at 0:27
Sam's answer is just right! – HipsterMathematician Sep 21 '12 at 0:28
Even the stripped-down MathJax version of $\TeX$ that we use here is fairly sophisticated: I changed ~$B$ and $\text{~} B$ to $\sim B$, coded as \sim B. – Michael Hardy Sep 21 '12 at 2:25
up vote 8 down vote accepted

Conditioning on an event reduces your probability space precisely to outcomes that satisfy the event in question, but not before "renormalizing" by that events probability to retain a probability function. In this case, you are conditioning on $A$. First, notice that $P(A\cap B)+P(A\cap B^c)=P(A)$ and by definition of conditional probability:

$$P(B|A) = \frac{P(B\cap A)}{P(A)}=\frac{P(A)-P(B^c\cap A)}{P(A)}=1-P(B^c|A)$$

In other words, if you keep $A$ fixed, then $P(\cdot |A)$ represents a perfectly good probability function on events.

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