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

\[p(A\cap B)= p(A|B) \cdot p(B)=p(B|A) \cdot p(A).\]


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References

Bibliography

  1. Bosch, Karl: "Elementare Einf├╝hrung in die Wahrscheinlichkeitsrechnung", vieweg Studium, 1995, 6th Edition