Conditional probability relation

Question

I'm trying to think about it graphically and analytically, but I can't seem to figure out why is the relation$P(A|B) = 1 - P(\overline{A} | B)$ true.

Is there an intuitive way of looking at it?

Answer 1

Assume that event B occurs.

Then, either event A occurs, or it does not occur. So, the sum of the probabilities of these two mutually exclusive situations must equal $(1)$.

The probability of the first situation occurring is $p(A|B).$

The probability of the second situation occurring is $p(\overline{A}|B).$

Answer 2

We have $$ \mathbb P(A|B) = \frac {\mathbb P(A\cap B)}{\mathbb P(B)} = \frac {\mathbb P(B\setminus (B\setminus A))}{\mathbb P(B)} = \frac {\mathbb P(B)- \mathbb P(B\setminus A)}{\mathbb P(B)} = 1 - \frac {\mathbb P(B\cap A^c)}{\mathbb P(B)} = 1- \mathbb P(A^c|B). $$

Intuitively, function $\mathbb P_B(A) = \mathbb P(A|B)$ defines probability measure, so naturally we have $\mathbb P_B(A) = 1 - \mathbb P_B(A^c)$. The fact that $\mathbb P_B$ is probability measure for any measurable $B$ such that $\mathbb P(B) > 0$ is a standard excercise.

Answer 3

$$P(A|B)+P(A^{\complement}\mid B)=P(A\cup A^{\complement}\mid B)=P(\Omega\mid B)=1$$

The map prescribed by: $$A\mapsto P(A\mid B)$$ is a probability measure.

It arises if the question is asked: "what are the probabilities of the events if it is for sure already that event $B$ occurs?"

Then naïvely we could go for the map prescribed by $A\mapsto P(A\cap B)$ but this is unfortunately not in general a probability measure because $P(\Omega\cap B)=P(B)$ so might take a value that is smaller than $1$.

This is repaired by division with $P(B)$ and makes us arrive at $$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$ (provided that $P(B)>0$)

It is straightforward to prove that this created map on events is indeed a probability measure.