There are 6 cards numbered 1 to 6 , one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two cards drawn. Then P(X > 3) is:

$\frac{14}{15}$

The correct answer is Option (1) → $\frac{14}{15}$

$\text{Total number of outcomes }=\frac{6!}{2!4!}=15.$

$X>3 \text{ except when } X=3.$

$X=3 \text{ only for } (1,2).$

$P(X=3)=\frac{1}{15}.$

$P(X>3)=1-\frac{1}{15}=\frac{14}{15}.$