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A shop is equipped with an alarm system which in case of burglary alerts police with probability $0.99$. During a night without a burglary, a false alarm is set off with probability $0.002$. The probability of a burglary on a given night is $0.0005$. An alarm has just gone off. What is the probability a burglarly is going on?

What I've got so far (not much):

$A:=$ "burglary alerts police" $\Rightarrow P(A)=0.99$

$B|C^{c}:=$"a false alarm is set off, given that it is a night without a burglarly" $\Rightarrow P(B|C^{c})=\frac{P(B \cap C^{c})}{P(C^{c})}=\frac{P(B \cap C^{c})}{1-P(C)}=\frac{P(B \cap C^{c})}{0.9995}=0.002\Rightarrow P(B \cap C^{c})=0.9995\times 0.002$

$C:=$"The probability of a burglary on a given night" $\Rightarrow P(C)=0.0005$

I this sense, I do not know how to define "Given that an alarm has just gone off, a burglarly is going on" set-theoretically.

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Let $A$ be the event in which a burglary is going on, $B$ the event in which the alarm is triggered. We then find:

$$P(B) = P(A) P(B | A) + P(\lnot A) P(B | \lnot A) = 0.0005 \cdot 0.99 + 0.9995 \cdot 0.002 = 0.002494$$

Using Bayes' theorem, we find:

$$P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.0005⋅0.99}{0.0005 \cdot 0.99 + 0.9995 \cdot 0.002} = \frac{0.000495}{0.002494} = 0.1985$$

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Monster math estimate: over $N$ days we expect about $.0005N$ robberies and alarms, and about $.002N$ false alarms. Thus the answer should be about $\frac {.0005}{.0005+.002}=.2$.

To do the calculation carefully:

There are two ways in which an alarm can go off: either there is a robbery and the alarm works as it should, or there is no robbery and the alarm sends a false signal.

The probability of the first case is $.0005\times .99=.000495$.

The probability of the second case is $(1-.0005)\times .002=.001999...$

Thus the total probability that an alarm goes off is the sum of these, $.0020494$.

Of this, the portion that is explained by a robbery is $.000495$ so the answer is the ratio $$\frac {.000495}{.0020494}=\boxed {.19848}$$

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