Total Probability of an Event

Theorem 12.10 -- Total Probability of an event. If $A_1,A_2,\ldots,A_n$ are mutually exclusive and exhaustive events (that is, they form a partition of the sample space $S$ -- see Section 12.3), and $B$ is ANY event in $S$, then $P(B)$ is called the total probability of the event $B$, and

Proof. Since $B$ is any event in $S$, and $A_1,\ldots,A_n$ partition $S$, the event $B$ itself splits into the pieces $(A_1\cap B),(A_2\cap B),\ldots,(A_n\cap B)$, whose union is exactly $B$: $B=(A_1\cap B)\cup(A_2\cap B)\cup\cdots\cup(A_n\cap B)$. Because $A_1,\ldots,A_n$ are mutually exclusive, so are these $n$ pieces of $B$, so by the additivity axiom $[P_2]$ (extended to $n$ events): $P(B)=P(A_1\cap B)+P(A_2\cap B)+\cdots+P(A_n\cap B)$. Finally, the Multiplication Theorem rewrites each term $P(A_i\cap B)=P(A_i)\cdot P(B/A_i)$, giving the stated formula.

How to recognise a Total Probability problem. The signal is a random experiment that happens in two natural stages: first, one of several mutually exclusive 'sources' or 'causes' $A_1,\ldots,A_n$ is determined (which urn is picked, which machine made the item, which plant produced the pipe); then, GIVEN that source, an event $B$ (drawing two red balls, an item being defective) has a probability that can differ from source to source. When the question asks for the OVERALL, unconditional probability of $B$ -- not yet 'given which source' -- Total Probability is exactly the weighted average $\sum P(A_i)P(B/A_i)$, weighted by how likely each source itself is.

Worked pattern (two-urn setting). Urn-I has 8 red and 4 blue balls, Urn-II has 5 red and 10 blue balls; one urn is chosen at random and two balls drawn from it. Selecting Urn-I ($A_1$) or Urn-II ($A_2$) are mutually exclusive and exhaustive, each with $P(A_1)=P(A_2)=\tfrac12$; the probability of drawing 2 red balls GIVEN each urn is computed separately by counting combinations within that urn, and the two contributions $P(A_1)P(B/A_1)$ and $P(A_2)P(B/A_2)$ are added. …