Probability of Fire? I have 
The Hotel has a fire alarm system which gives an alarm with a probability of 99.2% in case of a fire outbreak. Occasionally the system gives false alarms. According to the night porter this happens five times a year (corresponds to a probability of 5/365). The probability of a fire breaking out on a certain night is 0.05%. 
Question:
Someone spends a night at the Hotel and hears the fire alarm. What is the real probability of a fire?
Is my calculation true?
((360/365)* 0,005 * 0,992)
 A: The task is to deal with conditional probability, IMO.
Consider $4$ non-intersecting cases:
1. fire, alarm ($(5\cdot10^{-4})\cdot 0.992$)
2. fire, no alarm ($(5\cdot10^{-4})\cdot 0.008$)
3. no fire, alarm ($0.9995\cdot\frac{5}{365}$)
4. no fire, no alarm ($0.9995\cdot\frac{360}{365}$)
It is given that alarm rang. The probability of it is $(5\cdot10^{-4})\cdot 0.992+0.9995\cdot\frac{5}{365}$
and the probability of fire in this case is $\frac{(5\cdot10^{-4})\cdot 0.992}{(5\cdot10^{-4})\cdot 0.992+0.9995\cdot\frac{5}{365}}$ $=\frac{9052}{258927}$ $\approx 0.034959660$
A: Using Bayes’s Theorem, we have $$ P(\text{fire|alarm})=\frac{P(\text{fire}) P(\text{alarm|fire})}{P(\text{fire})P(\text{alarm|fire}) + P(\text{no fire}) P(\text{alarm|no fire})} \\ = \frac{0.05\% \cdot 99.2\%}{0.05\% \cdot 99.2\% + 99.95\% \cdot \frac{5}{365}} \\ =\frac{4.96}{4.96 +136.917} \\ \approx 0.035 $$
A: This is exactly the same answer provided before by Alexey Burdin, but with a longer explanation.
Let $F=\{0,1\}$ denote the absence ($0$) or the presence of fire ($1$). Let $A=\{0,1\}$ denote the absence ($0$) or the presence of alarm ($1$).
We know that:


*

*True positive probability: $P(A = 1 | F = 1) = 0.992.$

*False positive probability: $P(A = 1 | F = 0) = 5/365.$

*Fire probability: $P(F= 1) = 0.0005.$
You are looking for $P(F = 1 | A = 1)$, i.e. the probability of fire knowing that the alarm is ringing. By definition:
$$P(F = 1 | A = 1) = \frac{P(F = 1 ~\text{and}~ A = 1)}{P(A=1)}.$$
On the other hand, we know that:
$$P(A = 1 | F = 1) = 0.992 = \frac{P(F = 1 ~\text{and}~ A = 1)}{P(F=1)} = \frac{P(F = 1 ~\text{and}~ A = 1)}{0.0005}.$$
Therefore:
$$P(F = 1 ~\text{and}~ A = 1) = 0.992 \cdot 0.0005,$$
and hence:
$$P(F = 1 | A = 1) =  \frac{0.992 \cdot 0.0005}{P(A=1)}.$$
Now, applying the total probability law, we get:
$$P(A = 1) = P(A= 1| F=0)P(F=0) + P(A=1 | F= 1)P(F=1).$$
Notice that $P(F=1) = 1-P(F=0) = 0.9995.$ Hence:
$$P(A = 1) = (5/365)\cdot 0.9995 + 0.992 \cdot 0.0005$$.
Finally:
$$P(F = 1 | A = 1) =  \frac{0.992 \cdot 0.0005}{(5/365)\cdot 0.9995 + 0.992 \cdot 0.0005} \simeq 0.3496.$$
Hence, the probability you are looking for is $\simeq 35 \%$. 
