Question

$A$ and $B$ plays the game of cards with their individual pack of cards by cutting two cards each; the probability that their cuts contains exactly one common card is

• A. $\frac{50}{663}$
• B. $\frac{29}{663}$
• C. $\frac{23}{663}$
• D. None of these

Answer 1

The correct option is A $\frac{50}{663}$

Let $A$ draw two cards one $R$ and one other than $B$
and $B$ draw two cards one $R$ and one other than $A$
$n\left(S\right){=}^{52}{\text{C}}_2{\times }^{52}{\text{C}}_2$
Now we need $\sum P\left(R\right)$ where $P\left(R\right)$ given by
$P\left(R\right)=\frac{{}^1{\text{C}}_1\times 51}{{}^{52}{\text{C}}_2}\times \frac{50{\times }^1{\text{C}}_1}{{}^{52}{\text{C}}_2}$where $R$ can take $52$ values
$\therefore \sum P\left(R\right)=\frac{52\times 51\times 50}{52\times 52\times 51\times 51}\times 2\times 2$ $=\frac{50}{663}$