There is a fire sensor system in a building. In long series of experiments, it...

Step 1
Given Information :
In the case of a fire, the sensor system will be activated with probability 0.97
In a night without fire, the sensor system may be activated with probability 0.01.
The probability of a fire in this specific area is founded ${\displaystyle =\frac{2}{397}=0.005}$
First we will draw a probability contingency table with all the information provided:
$\begin{array}{|cccc|}\hline & \text{Sensor system is activate}& \text{Sensor system is not activate}& Total\\ \text{There is fire}& 0.97\cdot 0.005=0.0049& 0.005-0.0049=0.0001& \frac{2}{397}=0.005\\ \text{There is no fire}& 0.01\cdot (1-0.005)=0.00995& 0.995-0.00995=0.98505& 1-0.005=0.995\\ Total& 0.01485& 0.98515& 1\\ \hline\end{array}$
Step 2
Now, if The fire sensor system is activated. Find the probability that there was actually a fire
We will use the principle of conditional probability:
$$\begin{array}{|cc|}\hline & \text{Sensor system is activated}\\ \text{There is fire}& 0.97\cdot 0.005=0.0049\\ \text{There is no fire}& 0.01\cdot (1-0.005)=0.00995\\ Total& 0.01485\\ \hline\end{array}$$
${\displaystyle P\left(\text{There is fire}\mid \text{sensor system is activated}\right)=}$
${\displaystyle \frac{P(\text{There is fire}\cap \text{sensor system is activated})}{\text{P(sensor system is activated)}}}$
${\displaystyle P\left(\text{There is fire}\mid \text{sensor system is activated}\right)=\frac{0.0049}{0.01485}=0.329}$
probability that there was actually a fire ${\displaystyle =0.329}$