Probability — Class 12 Maths Solution

.: Let ${E_1}$ : ‘coin chosen is two headed’,

${E_2}$ : ‘coin chosen is biased’ and

${E_3}$ : ‘coin chosen is unbiased’,

$\Rightarrow$ $P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = P\left( {{E_3}} \right) = \cfrac{1}{3}$

Let $A$ : 'tossed coin shows up a head',

then '
$P\left( {A|{E_1}} \right) = 1,P\left( {A|{E_2}} \right) = \cfrac{{75}}{{100}} = \cfrac{3}{4}$

and $P\left( {A|{E_3}} \right) = \cfrac{1}{2}$

(The coin, which has head on both sides shows

head with surety and an unbiased coin shows head with probability $\cfrac{1}{2}$ ).

Hence, the required probability is

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

$= \cfrac{{1 \times \cfrac{1}{3}}}{{1 \times \cfrac{1}{3} + \cfrac{3}{4} \times \cfrac{1}{3} + \cfrac{1}{2} \times \cfrac{1}{3}}} = \cfrac{1}{{1 + \cfrac{3}{4} + \cfrac{1}{2}}} = \cfrac{1}{{\cfrac{{4 + 3 + 2}}{4}}} = \cfrac{4}{9}$