Probability — Class 12 Maths Solution

.: 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

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

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

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