A card from a pack of 52 cards 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.
A card from a pack of 52 cards 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.

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}