An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01,0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01,0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Solution:

Let \mathrm{E}_{1} represent the driver being a scooter driver, \mathrm{E}_{2} represent the driver being a car driver, and \mathrm{E}_{3} represent the driver being a truck driver. Assume that the person is involved in an accident.

=2000+4000+6000=12000 total number of drivers

Then \mathrm{P}\left(\mathrm{E}_{1}\right)=2000 / 12000=1 / 6

\mathrm{P}\left(\mathrm{E}_{2}\right)=4000 / 12000=1 / 3

P\left(E_{3}\right)=6000 / 12000=1 / 2

Because a headed coin has a head on both sides, it will display one.

Also \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\mathrm{P}( accident of a scooter driver )=0.01=1 / 100

And P\left(A \mid E_{2}\right)=P( accident of a car driver )=0.03=3 / 100

And P\left(A \mid E_{3}\right)=P( accident of a truck driver )=0.15=15 / 100=3 / 20

Because a headed coin has a head on both sides, it will display one. Given that the driver was involved in an accident, the probability that he was a scooter driver is \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right).

We have used Bayes’ theorem to arrive at our conclusion.

    \[ <!-- /wp:paragraph --> <!-- wp:paragraph --> \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)+\mathrm{P}\left(\mathrm{E}_{3}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{3}\right)} <!-- /wp:paragraph --> <!-- wp:paragraph --> \]

We may now retrieve the result by swapping the values.

=\frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{6} \cdot \frac{1}{100}+\frac{1}{3} \cdot \frac{3}{100}+\frac{1}{2} \cdot \frac{3}{20}}

=\frac{\frac{1}{600}}{\frac{1}{20}\left(\frac{1}{30}+\frac{1}{5}+\frac{3}{2}\right)}

=\frac{\frac{1}{30}}{\frac{52}{30}}=\frac{1}{52}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{1}{52}