# Proof

(related to Theorem: Law of Total Probability)

Because, by hypothesis, the events $$A_1,A_2,\ldots,A_n$$ are mutually exclusive and collectively exhaustive, it follows from the definition of probability. $1=p(\Omega)=p(A_1\cup A_2\cup\ldots \cup A_n)=\sum_{i=1}^n p(A_i).\quad\quad ( * )$

For any event $$B$$, the events $$(B\cap A_1), (B\cap A_2),\ldots, (B\cap A_n)$$ are also mutually exclusive. Therefore, we have

$p((B\cap A_1)\cup (B\cap A_2)\cup\ldots\cup (B\cap A_n))=\sum_{i=1}^n p(B\cap A_i).\quad\quad ( * * )$

By hypothesis, we have $$p(A_i) > 0$$ for $$i=1,2,\ldots,n$$ and it follows from the conditional probability formula:

$\begin{array}{rcll} p(B)&=&p(B\cap \Omega)&\text{definition of probability space}\\ &=&p(B\cap (A_1\cup A_2\cup\ldots \cup A_n))&\text{by }( * )\\ &=&p((B\cap A_1)\cup (B\cap A_2)\cup\ldots\cup (B\cap A_n))&\text{distributivity law for sets}\\ &=&\sum_{i=1}^n p(B\cap A_i)&\text{by }( * * )\\ &=&\sum_{i=1}^n p(B|A_i)p(A_i)&\text{conditional probability formula and }p(A_i) > 0\text{ for } i=1,2,\ldots,n.\\ \end{array}$

Github: ### References

#### Bibliography

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