# Bayes' rule for conditional probability

Problem statement:

Your neighbour has a broken alarm system. Whenever someone breaks into his house, the alarm goes off with a probability of 95%. However, during the last 8 weeks, the alarm went off 7 times, for no reason. The probability of someone breaking into the house is 0.0005. Suppose the alarm of your neighbour starts making noise, what is the probability of someone trying to break in?

So I named I the event of a break-in and A the event of the alarm working. This gives me probabilities $$P(A|I) = 0.95, \quad P(I) = 0.0005$$ and we are looking for $$P(I|A)$$. However, I have no idea what to do with the statement about the alarm going off 7 times in the last 8 weeks for no reason...

Using Bayes rule, the only probability which I don't know is $$P(A|\overline{I})$$, so I suspect this statement having to do something with this unknown probability.

Any suggestions? I hope no details went lost in translation, English is not my first language.

The last two weeks contains $$8 \cdot 7 = 56$$ nights, of which 7 times the alarm rang, therefore the probability of the alarm going off without anyone trying to break in must be approximately $$\frac 18$$
• @Student Using $7/8$ would make sense if the probabilities were per week rather than per day. The problem statement is ambiguous. And see math.stackexchange.com/questions/2279851/… Commented Jan 12, 2021 at 20:42