Question
At a reception desk, a visitor takes exactly one loan key from a tray. The visitor picks key A with probability \(2/5\) and key B with probability \(3/5\). Both outcomes occur sometimes, but the reception policy forbids taking both keys on the same visit. A supervisor says, “The two keys are stored in different lockers, so whether A is borrowed and whether B is borrowed are independent events.” Is that claim mathematically possible under the lending rule? Identify the relevant probabilities, and state the exceptional circumstance in which two mutually exclusive events could also be independent.
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Hint 1
Independence concerns event probabilities, not where the physical keys are stored.
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Solution
Let \(A\) be the event “key A is borrowed on this visit” and \(B\) be “key B is borrowed.” By the one-key policy, the events cannot occur together, so
\[
P(A\cap B)=0.
\]
Their individual probabilities are \(P(A)=2/5\) and \(P(B)=3/5\). Independence would require
\[
P(A\cap B)=P(A)P(B)=\frac25\frac35=\frac6{25},
\]
which is positive and contradicts exclusivity. Thus the supervisor's claim is false. The fact that the keys are stored separately does not change the process that selects exactly one key.
In general, disjoint events satisfy \(P(A\cap B)=0\). They are also independent exactly when \(P(A)P(B)=0\), which for probabilities means \(P(A)=0\) or \(P(B)=0\) (or both). Those zero-probability exceptions are excluded here because both keys are actually chosen with positive probability. This is a claim about events under the desk's loan rule, not a claim that two distinct physical mechanisms could never be independent in a different experiment.
Assumptions
Exactly one key is borrowed on each visit.
The two event probabilities are positive and sum to one.
Final Answer
No. \(P(A\text{ and }B)=0\) while \(P(A)P(B)=6/25>0\). Mutually exclusive events are independent only if at least one has probability zero.