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Can $\Pr(B|-A)$ be written as $1-\Pr(-B|-A)$?
Would be helpful if someone could confirm this.

Math Attack
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G2_2023
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    Have you tried a simple example? – copper.hat Apr 23 '23 at 20:53
  • Whatever you're given only changes the sample space. All the rules you learned still apply. In this case, if $A$ does not happen, it's still true that B either happens or does not, and those are still mutually exclusive. – David P Apr 23 '23 at 21:04

2 Answers2

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Conditional Probability is still a probability measure.

All the usual rules apply.

Such as "the sum of the probabilities of complementary events equals one".

$$\mathsf P(A\mid \neg B) + \mathsf P(\neg A\mid\neg B)=1$$

If you want to be sure, you can just use the definition of conditioning.

When $\mathsf P(\lnot B)>0$ then:

$$\begin{align} \mathsf P(A\mid \neg B) + \mathsf P(\neg A\mid\neg B) &= \dfrac{~\mathsf P(A\cap\neg B)~}{\mathsf P(\lnot B)}+\dfrac{~\mathsf P(\neg A\cap\neg B)~}{\mathsf P(\neg B)}&&\text{by definition}\\[1ex]&= \dfrac{~\mathsf P\big((A\cap\neg B)\cup(\neg A\cap\neg B)\big)~}{\mathsf P(\neg B)}&&\text{the events are disjoint}\\[1ex]&=\dfrac{~\mathsf P\big((A\cup\neg A)\cap\neg B\big)~}{\mathsf P(\neg B)}&&\text{distribution}\\[1ex]&=\dfrac{~\mathsf P(\neg B)~}{\mathsf P(\neg B)}\\[1ex]&=1 \end{align}$$ As was to be demonstrated.

$\blacksquare$

Graham Kemp
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Remember that $B | -A$ is the event that $B$ occurs given that $A$ does not. Call this event $E$. Then we have $P(-E) = 1 - P(E).$ Now back substitute.

Sean Roberson
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