Probabilistic question This is just a well known probability exercise, but I cannot find the answer. > Suppose there is an alarm for detecting burglars. Suppose there is a false alarm (not caused by a burglar and not caused by an earthquake) with probability $f$. Suppose an earthquake triggers the alarm with probability $\alpha_e$ and suppose that a burglar causes the alarm to ring with probability $\alpha_b$. Let $a$ denote the alarm (as binary variable), $e$ denote the earthquake and $b$ denote a burglar. Furthermore, $b$ and $e$ are independent. What is $P(a=0|b=1, e=0)$? The answer should be $(1-f)(1-\alpha_b)$ but I tried Bayes' theorem many times but could not find the answer. I tried: $$P(a=0|b=1,e=0)=\frac{P(a=0,b=1,e=0)}{P(b=1,e=0)}=\frac{P(a=0,b=1,e=0)}{P(b=1)P(e=0)}$$ But that simply does not work out.

Here is an attempt to recover the solution. Let $b$ be the event of a burgler, $e$ an earthquake, as in the question. Let $T_b\subset b$ be that there is a burgler and he trips the alarm, and let $T_e\subset e$ be defined similarly. Let $F$ be the event of a false alarm. In this notation, I interpret the problem to say $P(F)=f,P(T_b\mid b)=\alpha_b,P(T_e\mid e)=\alpha_e,$ and the probability of an alarm to be $P(F\cup T_b\cup T_e)$.

Then $P(\text{alarm} \mid b,e^c)=P(F^c\cap T_b^c\cap T_e^c \mid b,e^c)=P(F^c\cap T_b^c\cap T_e^c,b,e^c)/P(b,e^c)=P(F^c\cap (b-T_b)\cap e^c)/(P(b)P(e^c)).$

Now if you impose independence assumptions that were not spelled out (e.g. the false alarm is independent of the other events), you can rewrite the numerator to get $$ P(\text{alarm} \mid b,e^c)=P(F^c\cap (b-T_b)\cap e^c)/(P(b)P(e^c))=P(F^c)P(b-T_b)P(e^c)/(P(b)P(e^c))=(1-f)P(b-T_b\mid b)=(1-f)(1-\alpha_b). $$