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I'm very rusty here, so this is probably a lot simpler than I'm trying to make it. Taking an online stats course and encountered this question.... Systems A and B are candidate fault detection systems. If there is a fault, system A raises an alert with a probability of 0.9, and system B raises an alert with a probability of 0.95. If there is no fault, system A raises a (false) alarm with a probability of 0.2, and system B raises a (false) alarm with a probability of 0.1. The two systems operate independently.

Assume that the probability of a fault is 0.4. If both systems raise an alarm, what is the probability there is actually a fault?

So if we let:

  • A be the event system A raises an alarm
  • B be the event system B raises an alarm
  • F be the event there is actually a fault

I can gather that:

  • $P(A|F)=0.9$
  • $P(B|F)=0.95$
  • $P(A|F')=0.2$
  • $P(B|F')=0.1$

And I believe that $P(A \cap B|F) = P(A|F) \cdot P(B|F) = (0.9) \cdot (0.95) = 0.855$ since they're independent.

So I think this is a Bayes Theorem question where we are looking to find $P(F|A \cap B)$

But then I start with $P(F|A \cap B) = \frac{P(F) \cdot P(A \cap B|F)}{P(A \cap B)}$ and am not sure where to go with that $P(A \cap B)$ in the denominator. Is this an application of total probability to find $P(A \cap B)$ or am I way off?

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The calculation is going to depend on whether you interpret "The two systems operate independently" as conditional independence of fault detection $P(A \cap B \mid F) = P(A\mid F) \cdot P(B\mid F)$ and $P(A \cap B \mid F') = P(A\mid F') \cdot P(B\mid F')$ even though here they would together be inconsistent with unconditional independence $P(A \cap B ) = P(A) \cdot P(B)$. As another issue, in real life "operational independence" is not the same thing as predictive independence.

In answer to your actual question,

  • $P(F \mid A \cap B)=\dfrac{P(A \cap B \cap F)}{P(A \cap B)}$
  • $P(A \cap B) = P(A \cap B \cap F) + P(A \cap B \cap F')$
  • $P(A \cap B \cap F) = P(F) \cdot P(A \cap B\mid F)$
  • $P(A \cap B \cap F') = P(F) \cdot P(A \cap B\mid F')$

so

  • $P(A \cap B)= P(F) \cdot P(A \cap B\mid F)+ P(F') \cdot P(A \cap B\mid F')$

  • $P(F \mid A \cap B)=\dfrac{P(F) \cdot P(A \cap B\mid F)}{P(F) \cdot P(A \cap B\mid F)+ P(F') \cdot P(A \cap B\mid F')}$

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