Question

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.

Answer 1

Let $E_1,E_2,E$ are the events

$E_1:$ Lost card is diamond.

$E_2:$ Lost card is not a diamond.

$E:$ Two cards drawn from the remaining pack of diamonds

$\therefore P\left(E_1\right)=\frac{13}{52}=\frac{1}{4},P\left(E_2\right)=\frac{39}{52}=\frac{3}{4}$

Also, $P\left(E|E_1\right)=\frac{{}^{12}{C}_2}{{}^{51}{C}_2}=\frac{\frac{12\times 11}{1\times 2}}{\frac{51\times 50}{1\times 2}}=\frac{12\times 11}{51\times 50}$

and $P\left(E|E_2\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}=\frac{\frac{13\times 12}{1\times 2}}{\frac{51\times 50}{1\times 2}}=\frac{13\times 12}{51\times 50}$

Required probability $=P\left(E_1|E\right)$

$=\frac{P\left(E_1\right)P\left(E|E_1\right)}{P\left(E_1\right)P\left(E|E_1\right)+P\left(E_2\right)P\left(E|E_2\right)}$

$=\frac{\frac{1}{4}\times \frac{12\times 11}{51\times 50}}{\frac{1}{4}\times \frac{12\times 11}{51\times 50}+\frac{3}{4}\times \frac{13\times 12}{51\times 50}}$

$=\frac{12\times 11}{12\times 11+13\times 12\times 3}$

$=\frac{132}{132+468}$

$=\frac{132}{600}$

$=\frac{11}{50}$