 namenerk

2021-12-11

There is a fire sensor system in a building. In long series of experiments, it was found that the accuracy of the fire sensor system can be described as follows:
- 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 $\frac{2}{397}$ from the past data.
The fire sensor system is activated. Find the probability that there was actually a fire. vrangett

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 $=\frac{2}{397}=0.005$
First we will draw a probability contingency table with all the information provided:
$\begin{array}{|cccc|}\hline & \text{Sensor system is activate}& \text{Sensor system is not activate}& Total\\ \text{There is fire}& 0.97\cdot 0.005=0.0049& 0.005-0.0049=0.0001& \frac{2}{397}=0.005\\ \text{There is no fire}& 0.01\cdot \left(1-0.005\right)=0.00995& 0.995-0.00995=0.98505& 1-0.005=0.995\\ Total& 0.01485& 0.98515& 1\\ \hline\end{array}$
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:
$\begin{array}{|cc|}\hline & \text{Sensor system is activated}\\ \text{There is fire}& 0.97\cdot 0.005=0.0049\\ \text{There is no fire}& 0.01\cdot \left(1-0.005\right)=0.00995\\ Total& 0.01485\\ \hline\end{array}$
$P\left(\text{There is fire}\mid \text{sensor system is activated}\right)=$
$\frac{P\left(\text{There is fire}\cap \text{sensor system is activated}\right)}{\text{P(sensor system is activated)}}$
$P\left(\text{There is fire}\mid \text{sensor system is activated}\right)=\frac{0.0049}{0.01485}=0.329$
probability that there was actually a fire $=0.329$

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