$\mathrm{E}_1=$ Event that lost card is diamond,
$\mathrm{E}_2=$ Event that lost card is not diamond.
Thereare 13 diamond cards, out of a pack or 52 cards
$$
\mathrm{P}\left(\mathrm{E}_1\right)=\frac{{ }^{13} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1}=\frac{13}{52}=\frac{1}{4}
$$
There are 39 cards which are not diamond,
$$
\mathrm{P}\left(\mathrm{E}_2\right)=\frac{39}{52}=\frac{3}{4}
$$
(i) When one diamond card is lost, 12 diamond cards are left and in total 51 cards are left. Out of 12 cards 2 may be drawn in ${ }^{12} \mathrm{C}_2$ way.
$\therefore$ Probability of getting 2 diamond cards when one diamond card is lost
$$
\mathrm{P}\left(\mathrm{A} / \mathrm{E}_1\right)=\frac{{ }^{12} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}=\frac{12 \times 11}{51 \times 50}
$$
Where A denotes the lost card
When diamond card is not lost, there are 13 diamond cards. The probability of drawing 2 diamond cards
$$
=\frac{{ }^{13} \mathrm{C}_2}{{ }^{51} \mathrm{C}_2}=\frac{13 \times 12}{51 \times 50}
$$
Probability that the lost card is diamond
$$
\begin{aligned}
&=P\left(E_1 / A\right)=\frac{P\left(E_1\right) P\left(A / E_1\right)}{\left(P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)\right.} \\
&=\frac{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50}}{\frac{1}{4} \times \frac{12 \times 11}{51 \times 50}+\frac{3}{4} \times \frac{13}{51} \times \frac{12}{50}}=\frac{11}{50}
\end{aligned}
$$