# Proof

(related to Proposition: Probability of Joint Events)

We will first show that for any two events $$A$$ and $$B$$, the probability of the joint event $$A\cap B$$ is the product of the conditional probability $$p(A|B)$$ and the probability of $$B$$:

$p(A\cap B)= p(A|B) \cdot p(B).\quad\quad ( * )$

### Case $$p(B)=0$$

Because $$A\cap B\subseteq B$$, it follows from the probability of included event that

$0\le p(A\cap B)\le p(B)\le 0$

Therefore, $$( * )$$ trivially holds.

### Case $$p(B) > 0$$

$$( * )$$ follows immediately from the formula of conditional probability.

It remains to be shown that

$p(A\cap B)= p(B|A) \cdot p(A).\quad\quad ( * * )$

Because $$A\cap B=B\cap A$$, we have

$p(B\cap A)= p(A|B) \cdot p(B).$

For symmetry reasons, we get from $$( * )$$ and $$( * * )$$ the required result

$p(A\cap B)= p(A|B) \cdot p(B)=p(B|A) \cdot p(A).$

Github: ### References

#### Bibliography

1. Bosch, Karl: "Elementare Einführung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition