Probability — Class 12 Maths Solution

.: (A) Let ${E_1}:$ ‘coin comes up with a head' and
${E_2}:$ ‘coin comes up with a tail'

$\Rightarrow P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = \cfrac{1}{2}$

Let $E:$ A reports that a head appears',

$P\left( {E|{E_1}} \right) = P$

(head comes up and $A$ speaks truth) $= \cfrac{4}{5}$

and $P\left( {E|{E_2}} \right) = P$

(tail comes up and $A$ tell a lie) $= 1 - \cfrac{4}{5} = \cfrac{1}{5}$

Required probability $= P\left( {{E_1}/E} \right)$

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

$= \cfrac{{\cfrac{4}{5} \times \cfrac{1}{2}}}{{\cfrac{4}{5} \times \cfrac{1}{2} + \cfrac{1}{5} \times \cfrac{1}{2}}} = \cfrac{4}{{4 + 1}} = \cfrac{4}{5}$