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Here's the question: Bob's burglary alarm is on. If there was a burglary, the alarm goes off with probability 50%. However, on any given day, there is a 1% chance the alarm is triggered by a dog. The burglary rate for the area is 1 burglary in $10^4$ days. What is the probability that Bob has been burglarized?

My solution is to use Baye's Rule:

$P(burg|alarm) = \frac{P(alarm|burg)*P(burg)}{P(alarm)}=\frac{P(alarm|burg)*P(burg)}{P(alarm|burg)+P(alarm|noBurg)} = \frac{\frac{1}{2}*\frac{1}{10^4}}{\frac{1}{100}+\frac{1}{2}}$

but this is apparently incorrect. Could someone please help me understand where my reasoning went wrong?

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1 Answer 1

up vote 2 down vote accepted

Your denominator is wrong. The correct formula for denominator: $P(alarm)=P(alarm|burg)*P(burg)+P(alarm|noBurg)*P(noBurg)$

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