Question

A card from a pack of $52$ cards is lost. From the remaining cards, two cards are drawn and are found to be spades. The probability that the missing card is also a spade is $\frac{a}{b}$ then the value of $a+b$ is

Answer 1

Let $A_1$ be the event that missing card is spade and $A_2$ be the event that missing card is non spade. Then,
$P\left(A_1\right)=\frac{1}{4}$, $P\left(A_2\right)=\frac{3}{4}$
Let $A$ be the event that $2$ spade cards are drawn from the remaining cards. Then,
$P\left(\frac{A}{A_1}\right)=\frac{{}^{12}{C}_2}{{}^{51}{C}_2}$ and $P\left(\frac{A}{A_2}\right)=\frac{{}^{13}{C}_2}{{}^{51}{C}_2}$
$\Rightarrow P\left(A\right)=P\left(A_1\right).P\left(\frac{A}{A_1}\right)+P\left(A_2\right).P\left(\frac{A}{A_2}\right)$
$=\frac{1}{4}\times \frac{{}^{12}{C}_2}{{}^{51}{C}_2}+\frac{3}{4}\times \frac{{}^{13}{C}_2}{{}^{51}{C}_2}$
$=\frac{1}{4\times {}^{51}{C}_2}[{}^{12}{C}_2+3{\times }^{13}{C}_2]$

Now,
$P\left(\frac{A_1}{A}\right)=\frac{P\left(A_1\right).P\left(\frac{A}{A_1}\right)}{P\left(A\right)}$
$=\frac{\frac{1}{4}\times \frac{{}^{12}{C}_2}{{}^{51}{C}_2}}{\frac{1}{4\times {}^{51}{C}_2}[{}^{12}{C}_2+3{\times }^{13}{C}_2]}$
$=\frac{11}{50}$
$\Rightarrow \frac{11}{50}=\frac{a}{b}$
$\Rightarrow a+b=11+50=61$