Question

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.

Answer 1

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,

$P\left(E1\right)=13/52=1/4$

$P\left(E2\right)=39/52=3/4$

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 $P\left(A\right|E1)$.

Also $P\left(A\right|E1)=12C2/51C2$

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

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

$P\left(A\right|E1)=12!/(10!.2!)x(49!.2!)/51!=(12\times 11)/(51x50)$

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

Also $P\left(A\right|E2)=13C2/51C2=(13x12)/(51x50)$

By putting the above values into the formula we get;
$P\left(E_1/A\right)=\frac{P\left(E_1\right).P\left(A/E_1\right)}{P\left(E_1\right).P\left(A/E_1\right)+P\left(E_2\right).P\left(A/E_2\right)}$
$$\begin{gathered} =\frac{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{12\times 11}{51\times 50}}}{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{12\times 11}{51\times 50}}+{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{13\times 12}{51\times 50}}} \\ =\frac{\frac{1}{4}\times \frac{12}{51\times 50}}{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{12}{51\times 50}}\left(11+3\times 13\right)} \\ =11/\left(11+39\right) \\ =11/50 \end{gathered}$$
Hence the required probability is $11/50$