Solution:
Let E1 represent the probability that the drawn card is a diamond, E2 represent the probability that the drawn card is not a diamond, and $\mathrm{A}$ represent the probability that the card is lost.
As we all know, there are 13 diamond cards and 39 non-diamond cards in a deck of 52 cards.
Then $\mathrm{P}\left(\mathrm{E}_{1}\right)=13 / 52$ and $\mathrm{P}\left(\mathrm{E}_{2}\right)=39 / 52$
When a diamond card is lost, there are now 12 diamond cards remaining out of a total of 51. In ${}^{12} \mathrm{C}_{2}$ ways, two diamond cards can be selected from a deck of 12 diamond cards.
In the same way, two diamond cards can be picked from a total of 51 cards in ${}^{51} \mathrm{C}_{2}$ ways.
When one diamond card is lost, the chance of obtaining two cards is $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)$.
Also $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)={ }^{12} \mathrm{C}_{2} /{ }^{51} \mathrm{C}_{2}$
Also $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)={ }^{12} \mathrm{C}_{2} /{ }^{51} \mathrm{C}_{2}$
$=\frac{12 !}{2 ! \times 10 !} \times \frac{2 ! \times 49 !}{51 !}$
$=\frac{12 \times 11 \times 10 !}{2 \times 1 \times 10 !} \times \frac{2 \times 1 \times 49 !}{51 \times 50 \times 49 !}$
$=\frac{12 \times 11}{51 \times 50}=\frac{22}{425}$
When no diamond card is lost, there are 13 diamond cards remaining out of a total of 51.
In ${}^{13} \mathrm{C}_{2}$ ways, two diamond cards can be selected from a deck of 13 diamond cards.
In the same way, two diamond cards can be picked from a total of 51 cards in ${}^{51} \mathrm{C}_{2}$ ways.
When a non-diamond card is lost, the probability of obtaining two cards is $\mathrm{P}$.
$\left(\mathrm{A} \mid \mathrm{E}_{2}\right)$.
Also $P\left(A \mid E_{2}\right)={ }^{13} C_{2} /{ }^{51} C_{2}$
$=\frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !}$
$=\frac{13 \times 12 \times 11 !}{2 \times 1 \times 10 !} \times \frac{2 \times 1 \times 49 !}{51 \times 50 \times 49 !}$
$=\frac{13 \times 12}{51 \times 50}=\frac{26}{425}$
Given that the card is lost, the probability that the lost card is diamond is $P\left(E_{1} \mid A\right)$.
Using Bayes’ theorem, we may deduce:
$\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_{2}\right)}$
We may now retrieve the result by swapping the values.
$=\frac{\frac{1}{4} \cdot \frac{22}{425}}{\frac{1}{4} \cdot \frac{22}{425}+\frac{3}{4} \cdot \frac{26}{425}}$
$=\frac{\frac{1}{425} \cdot \frac{22}{4}}{\frac{1}{425}\left(\frac{22}{4}+\frac{26 \times 3}{4}\right)}$
$=\frac{\frac{11}{2}}{\frac{100}{4}}=\frac{11}{50}$
$\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{A}\right)=\frac{11}{50}$