Question
Two rehearsal rooms share a ventilation controller, which is secretly either in poor mode or good mode, each with prior probability 1/2. Conditional on the true mode, the rooms independently develop excessive condensation during a rehearsal: each has probability 3/4 in poor mode and 1/4 in good mode. A supervisor does not see the controller mode. Instead the supervisor sees a red warning lamp, which lights with probability 3/4 in poor mode and 1/4 in good mode. Given the true mode, the lamp and both rooms' condensation events are mutually independent.
After the lamp lights red, what is the probability both rooms develop condensation? Are the two rooms' condensation events independent conditional on that red warning? Justify your conclusion quantitatively.
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Hint 1
Use Bayes’ rule to update the probability that the hidden ventilation mode is poor after seeing red.
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Solution
Let H denote poor ventilation, R a red lamp, and A,B condensation in rooms A and B. By Bayes’ rule, \(P(H\mid R)=\frac{P(R\mid H)P(H)}{P(R\mid H)P(H)+P(R\mid H^c)P(H^c)}=\frac{(\frac34\cdot\frac12)}{(\frac34\cdot\frac12+\frac14\cdot\frac12)}=\frac34\); \(P(H^c\mid R)=\frac14\).
Within each true mode, A and B are independent and remain independent of R. Consequently, \(P(A\cap B\mid H,R)=(\frac34)^2=\frac9{16}\), and \(P(A\cap B\mid H^c,R)=(\frac14)^2=\frac1{16}\). Averaging with the posterior weights gives \(P(A\cap B\mid R)=(\frac34)(\frac9{16})+(\frac14)(\frac1{16})=\frac{27+1}{64}=\frac{28}{64}=\frac7{16}\).
Each room's marginal after red is \(P(A\mid R)=P(B\mid R)=(\frac34)(\frac34)+(\frac14)(\frac14)=\frac{9+1}{16}=\frac{10}{16}=\frac58\). If A and B were conditionally independent given red, their joint probability would be \((\frac58)^2=\frac{25}{64}\), but the actual joint probability is \(\frac{28}{64}\). Their conditional covariance as binary indicators is \(\frac{28}{64}-\frac{25}{64}=\frac3{64}>0\). Knowing the actual ventilation mode would remove this shared-source dependence; seeing a noisy proxy does not, because the posterior still mixes two different condensation probabilities.
Assumptions
Prior probabilities of the two ventilation modes are each 1/2.
Conditional on a mode, both room indicators and the lamp are mutually independent with the stated mode-dependent Bernoulli probabilities.
The supervisor observes red but neither the mode nor either room’s outcome.
Final Answer
\(P(\text{both rooms condense}\mid\text{red})=\frac7{16}=0.4375\). Each room has conditional probability 5/8, whose product is \(\frac{25}{64}\ne\frac7{16}\); dependence remains, with conditional covariance 3/64.