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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 E1â€‹,E2â€‹, and E3â€‹ 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 Bayeâ€™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 Bayeâ€™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.

Complete step-by-step answer:

Let E1â€‹,E2â€‹, and E3â€‹ 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 Bayeâ€™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 Bayeâ€™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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