A card from a pack of 52 cards is lost.

Let E1 and E2 be the respective events of choosing a diamond card and a card which is not diamond.

Let A denote the lost card.

Out of 52 cards, 13 cards are diamond and 39 cards are not diamond.

$\therefore P\left(E_{1}\right)=\frac{13}{52}=\frac{1}{4}$

$\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{39}{52}=\frac{3}{4}$

When one diamond card is lost, there are 12 diamond cards out of 51 cards.

Two cards can be drawn out of 12 diamond cards in ${ }^{12} \mathrm{C}_{2}$ ways.

Similarly, 2 diamond cards can be drawn out of 51 cards in ${ }^{51} \mathrm{C}_{2}$ ways. The probability of getting two cards, when one diamond card is lost, is given by $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)$.

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)=\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{12 !}{2 ! \times 10 !} \times \frac{21 \times 49 !}{51 !}=\frac{11 \times 12}{50 \times 51}=\frac{22}{425}$

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.

Two cards can be drawn out of 13 diamond cards in ${ }^{13} \mathrm{C}_{2}$ ways whereas 2 cards can be drawn out of 51 cards in ${ }^{51} \mathrm{C}_{2}$ ways.

The probability of getting two cards, when one card is lost which is not diamond, is given by $P\left(A \mid E_{2}\right)$.

$\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)=\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !}=\frac{12 \times 13}{50 \times 51}=\frac{26}{425}$

The probability that the lost card is diamond is given by $P\left(E_{1} \mid A\right)$.

By using Bayes’ theorem, we obtain

$P\left(E_{1} \mid A\right)=\frac{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)}{P\left(E_{1}\right) \cdot P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \cdot P\left(A \mid E_{2}\right)}$

$=\frac{\frac{1}{4} \cdot \frac{22}{425}}{\frac{1}{4} \cdot \frac{22}{425}+\frac{3}{4} \cdot \frac{26}{425}}$

$=\frac{\frac{1}{425}\left(\frac{22}{4}\right)}{\frac{1}{425}\left(\frac{22}{4}+\frac{26 \times 3}{4}\right)}$

$=\frac{11}{2}$

$=\frac{11}{50}$