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

.: Let ${E_1}$ : `Insured person is a scooter driver',

${E_2}$ : ‘Insured person is a car driver’ and

${E_3}$ : `Insured person is truck driver,

$\Rightarrow$ $P\left( {{E_1}} \right) = \cfrac{{2000}}{{2000 + 4000 + 6000}} = \cfrac{1}{6}$

$P\left( {{E_2}} \right) = \cfrac{{4000}}{{2000 + 4000 + 6000}} = \cfrac{1}{3}$

and
$P\left( {{E_3}} \right) = \cfrac{{6000}}{{2000 + 4000 + 6000}} = \cfrac{1}{2}$

Let ${\rm{A}}:$ Insured person meets with an accident',

then $P\left( {A|{E_1}} \right) = 0.01 = \cfrac{1}{{100}},P\left( {A|{E_2}} \right) = 0.03 = \cfrac{3}{{100}},$

and
$P\left( {A|{E_3}} \right) = 0.15 = \cfrac{{15}}{{100}}$

Required probability

$= P\left( {{E_1}|A} \right) = \cfrac{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right)}}{{P\left( {A|{E_1}} \right)P\left( {{E_1}} \right) + P\left( {A|{E_2}} \right)P\left( {{E_2}} \right) + P\left( {A|{E_3}} \right)P\left( {{E_3}} \right)}}$

$= \cfrac{{\cfrac{1}{{100}} \times \cfrac{1}{6}}}{{\cfrac{1}{{100}} \times \cfrac{1}{6} + \cfrac{3}{{100}} \times \cfrac{1}{3} + \cfrac{{15}}{{100}} \times \cfrac{1}{2}}}$

$= \cfrac{{\cfrac{1}{6}}}{{\cfrac{1}{6} + 1 + \cfrac{{15}}{2}}} = \cfrac{1}{{1 + 6 + 45}} = \cfrac{1}{{52}}$ .