Probability of not $A$ and knowing that $B$ has not happened.

Question

I'm clear that $$P(A\mid B) = \frac{P(A∩B)}{P(B)}$$

But how can you calculate the following: $P(\bar A \mid \bar B )$?

Thanks

Answer 1

By using the same formula: $$ P(\bar A\mid \bar B)=\frac{P(\bar A\cap\bar B)}{P(\bar B)} $$ Note that the formula doesn't care what your two events are called, so if you have two events that you have decided to call $わ$ and $\dagger$, then we know that $$ P(わ\mid \dagger)=\frac{P(わ\cap\dagger)}{P(\dagger)} $$

Answer 2

Assuming that $P(B),P(\bar B)>0$ (just to avoid trivial cases) you have that for any event $A$\begin{align}P(A)&=P(A\mid B)P(B)+P(A\mid \bar B)P(\bar B)=P(A\mid B)P(B)+(1-P(\bar A\mid \bar B))P(\bar B)\\[0.2cm]\implies P(\bar A\mid \bar B)&= 1-\frac{P(A)-P(A\mid B)P(B)}{P(\bar B)} \end{align} which can be further simplified/ manipulated.