Question
A factory currently approves a batch only if it passes inspections A and B. Management proposes a stricter rule: a batch must pass A, B and a new inspection C. All three results may be dependent; you are given no pass probabilities or information about how the tests are related.
Can adding C increase the probability of approval? Must the probability strictly decrease? State the exact condition under which the approval probability remains unchanged, and give an example showing that equality is possible even when C sometimes fails.
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Hint 1
Think of each approval rule as a set of batches. Every batch satisfying the new rule also satisfies the old one.
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Solution
Let the letters also denote the events that their corresponding inspections pass. The original rule approves on \(O=A\cap B\), whereas the revised rule approves on \(N=A\cap B\cap C\). Every outcome in \(N\) is also in \(O\), so \(N\subseteq O\), which proves
\[ P(N)\le P(O) \]
with no independence assumption. More precisely, \(O\) is the disjoint union of \(N\) and \(A\cap B\cap C^c\). Therefore
\[ P(O)-P(N)=P(A\cap B\cap C^c). \]
It follows that equality holds if and only if the extra test never rejects a batch that would otherwise be approved, except on a probability-zero set: \(P(A\cap B\cap C^c)=0\). A strict decrease occurs exactly when that event has positive probability.
For example, toss a fair coin to determine A; let B always pass, and define C to pass exactly when A does. Then \(P(C\text{ fails})=1/2\), but both old and new rules approve precisely on heads, with probability \(1/2\). The result does not depend on whether individual test failures are common or rare; it follows from event inclusion alone.
Assumptions
\(A\), \(B\) and \(C\) denote pass events under the same underlying distribution of batches.
The addition of C does not change how A or B is administered or the tested population.
No independence, conditional independence or particular pass rate is assumed.
Final Answer
The new approval probability cannot exceed the old one: \(P(A\cap B\cap C)\le P(A\cap B)\). Equality holds exactly when \(P(A\cap B\cap C^c)=0\). A strict decrease is not guaranteed.