Questions & Answers

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of 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?

Answer Verified Verified
Hint: First you have to find the probability of driver that means either driver is scooter driver or car driver or truck driver. Then you have to find the probability of a driver with an accident that means either truck driver met accident or car driver or scooter driver. And then apply Bayes theorem to get an answer.

Complete step-by-step answer:
Let E1​,E2​, and E3​ be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let A be the event that the person meets with an accident.
There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers
Total number of drivers = 2000 +4000+ 6000 =12000
$P\left( {{E_1}} \right) = P\left( {{\text{Driver is a scooter driver}}} \right) = \dfrac{{2000}}{{12000}} = \dfrac{1}{6}$
$P\left( {{E_2}} \right) = P\left( {{\text{Driver is a car driver}}} \right) = \dfrac{{4000}}{{12000}} = \dfrac{1}{3}$
$P\left( {{E_3}} \right) = P\left( {{\text{Driver is a truck driver}}} \right) = \dfrac{{6000}}{{12000}} = \dfrac{1}{2}$
$P\left( {\dfrac{A}{{{E_1}}}} \right) = P\left( {{\text{Scooter driver with an accident}}} \right) = 0.01 = \dfrac{1}{{100}}$
$P\left( {\dfrac{A}{{{E_2}}}} \right) = P\left( {{\text{Car driver with an accident}}} \right) = 0.03 = \dfrac{3}{{100}}$
$P\left( {\dfrac{A}{{{E_3}}}} \right) = P\left( {{\text{Truck driver with an accident}}} \right) = 0.15 = \dfrac{{15}}{{100}}$
The probability that driver is a scooter driver , given that he met with an accident , is given by$P\left( {\dfrac{{{E_1}}}{A}} \right)$
By using Baye’s theorem , we obtain
$P\left( {\dfrac{{{E_1}}}{A}} \right) = \dfrac{{P\left( {{E_1}} \right).P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{P\left( {{E_1}} \right).P\left( {\dfrac{A}{{{E_1}}}} \right) + P\left( {{E_2}} \right).P\left( {\dfrac{A}{{{E_2}}}} \right) + P\left( {{E_3}} \right).P\left( {\dfrac{A}{{{E_3}}}} \right)}}$
$ = \dfrac{{\dfrac{1}{6}.\dfrac{1}{{100}}}}{{\dfrac{1}{6}.\dfrac{1}{{100}} + \dfrac{1}{3}.\dfrac{3}{{100}} + \dfrac{1}{2}.\dfrac{{15}}{{100}}}}$
$ = \dfrac{{\dfrac{1}{6}.\dfrac{1}{{100}}}}{{\dfrac{1}{{100}}\left( {\dfrac{1}{6} + 1 + \dfrac{{15}}{2}} \right)}} = \dfrac{{\dfrac{1}{6}}}{{\dfrac{{52}}{6}}} = \dfrac{1}{6} \times \dfrac{{12}}{{104}} = \dfrac{1}{{52}} = 0.019$

Note: Whenever you get this type of question the key concept of solving is you should have knowledge of Baye’s theorem and you have to understand $P\left( {\dfrac{A}{{{E_1}}}} \right)$ means probability of event A after completion of event ${{\text{E}}_1}$. And also understand that probability means number of favorable outcomes divided by number of total outcomes.
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