Question

$A$ draws a card from a pack of $n$ cards marked $1,2,...,n.$ The card is replaced in the pack and $B$ draws a card. Then the probability that $A$ draws a higher card than $B$ is

• A. $\frac{(n+1)}{2n}$
• B. $\frac{1}{2}$
• C. $\frac{(n-1)}{2n}$
• D. $\frac{(n-1)}{n}$

Answer 1

The correct option is C $\frac{(n-1)}{2n}$

If $A$ draws card higher than $B,$ then number of favourable cases is $(n-1)+(n-2)+\cdots +3+2+1$
(as when $B$ draws card number $1,$ then $A$ can draw any card from $2$ to $n$ and so on).
Therefore, the required probability is
$\frac{\frac{n(n-1)}{2}}{n^2}=\frac{n-1}{2n}$