Acard from a pack of 52 card is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

.: 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

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

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

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