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

Let,

E1 be the event that the drawn card is a diamond

E2 be the event that the drawn card is not a diamond and

A be the event that the card is lost

As we know, in the 52 deck of cards, 13 cards are diamond and 39 cards are not diamond.

So,

P (E1) = 13/52 = 1/4

P (E2) = 39/52 = 3/4

After losing one diamond car, there are 12 diamond cards remaining out of 51 cards.

Two diamond cards can be drawn out of 12 diamond cards in 12C2 ways.

Similarly, two diamond cards can be drawn out of total 51 cards in 51C2 ways.

Thus, the probability of getting two cards, when one diamond card is lost, is P (A|E1).

Also P (A|E1) =12C2 / 51C2

12C2 = 12!/(10!.2!)

51C2 = 51!/(49!.2!)

P (A|E1) = 12!/(10!.2!) x (49!.2!)/51! = (12×11)/(51x 50)

Now, probability of getting two cards, when card is lost which is not diamond, is P (A|E2).

Also P (A|E2) =13C2 / 51C2 = (13 x 12)/(51 x 50)

By putting the above values into the formula we get;

\begin{aligned} \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{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} \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 \times 12}{51 \times 50}} \\ &=\frac{\frac{1}{4} \times \frac{12}{51 \times 50}}{\frac{1}{4} \times \frac{12}{51 \times 50}(11+3 \times 13)} \end{aligned}

= 11/(11+39) = 11/50

Hence, the required probability is 11/50.