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?
