Probability — Class 12 Maths Solution

.: Let ${E_1}:$ lost card is a diamond', ${E_2}:$

'lost card is not a diamond',

$\Rightarrow P\left( {{E_1}} \right) = \cfrac{{13}}{{52}} = \cfrac{1}{4}$

and $P\left( {{E_2}} \right) = \cfrac{{39}}{{52}} = \cfrac{3}{4}$

Let $A$ : two cards drawn from the remaining pack are diamonds',

then, $P\left( {A|{E_1}} \right) = \cfrac{{^{12}{C_2}}}{{^{51}{C_2}}} = \cfrac{{\cfrac{{12 \times 11}}{{1 \times 2}}}}{{\cfrac{{51 \times 50}}{{1 \times 2}}}} = \cfrac{{12 \times 11}}{{51 \times 50}}$

and $P\left( {A|{E_2}} \right) = \cfrac{{^{13}{C_2}}}{{^{51}{C_2}}} = \cfrac{{13 \times 12}}{{51 \times 50}}$

Required probability

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

$= \cfrac{{\cfrac{{12}}{{51}} \times \cfrac{{11}}{{50}} \times \cfrac{1}{4}}}{{\cfrac{{12 \times 11}}{{51 \times 50}} \times \cfrac{1}{4} + \cfrac{{13 \times 12}}{{51 \times 50}} \times \cfrac{3}{4}}} = \cfrac{{12 \times 11}}{{12 \times 11 + 13 \times 12 \times 3}}$

$= \cfrac{{132}}{{132 + 468}} = \cfrac{{132}}{{600}} = \cfrac{{11}}{{50}}$