There is a fire sensor system in a building. In long series of experim

namenerk 2021-12-11 Answered
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 \(\displaystyle{\frac{{{2}}}{{{397}}}}\) from the past data.
The fire sensor system is activated. Find the probability that there was actually a fire.

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Expert Answer

vrangett
Answered 2021-12-12 Author has 4182 answers

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:
\(\begin{array}{|c|c|} \hline &\text{Sensor system is activate}&\text{Sensor system is not activate}&Total\\ \hline \text{There is fire}&0.97 \cdot 0.005=0.0049&0.005-0.0049=0.0001&\frac{2}{397}=0.005\\ \hline \text{There is no fire}&0.01 \cdot (1-0.005)=0.00995&0.995-0.00995=0.98505&1-0.005=0.995\\ \hline Total&0.01485&0.98515&1\\ \hline \end{array}\)
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:
\[\begin{array}{|c|c|} \hline &\text{Sensor system is activated}\\ \hline \text{There is fire}&0.97 \cdot 0.005=0.0049\\ \hline \text{There is no fire}&0.01 \cdot (1-0.005)=0.00995\\ \hline Total&0.01485\\ \hline \end{array}\]
\(\displaystyle{P}{\left(\text{There is fire}{\mid}\text{sensor system is activated}\right)}=\)
\(\displaystyle{\frac{{{P}{\left(\text{There is fire}\cap\text{sensor system is activated}\right)}}}{{\text{P(sensor system is activated)}}}}\)
\(\displaystyle{P}{\left(\text{There is fire}{\mid}\text{sensor system is activated}\right)}={\frac{{{0.0049}}}{{{0.01485}}}}={0.329}\)
probability that there was actually a fire \(\displaystyle={0.329}\)

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