Q.A factory has two Machines I and II. Machine I produces 60% of the items and Machine II produces 40% of the items of the total output. Further, 2% of the items produced by Machine I are defective whereas 4% produced by Machine II are defective. If an item is drawn at random, what is the probability that it is defective?

Theorem 12.10 (Total Probability of an Event). Suppose $A_1,A_2,\ldots,A_n$ are mutually exclusive AND exhaustive events (a partition of the sample space $S$), and $B$ is any event in $S$. Then

Why it works. Since $A_1,\ldots,A_n$ partition $S$, the event $B$ splits into the mutually exclusive pieces $(A_1\cap B),(A_2\cap B),\ldots,(A_n\cap B)$, whose union is $B$ itself. Adding their probabilities (axiom $[P_2]$) gives $P(B)=\sum_i P(A_i\cap B)$, and the Multiplication Theorem rewrites each term as $P(A_i)P(B/A_i)$.

When to reach for it. Total probability is the tool whenever an event $B$'s chance genuinely depends on WHICH of several disjoint 'causes' or 'sources' occurred first -- which urn was picked, which machine made the item, which agency the car came from -- and you are asked for $B$'s overall (unconditional) probability, not yet 'given which cause'. It is exactly a probability-weighted average of the conditional probabilities $P(B/A_i)$, weighted by how likely each cause $A_i$ is. This same partition-and-weight structure is the setup that Bayes' Theorem then reverses.