The number on the first card needs to be One higher than that on the second card, so possibilities are :
$\begin{array}{c} \begin{array}{cc} 1^{\text{st}} \text{ card} & 2^{\text{nd}} \text{ card}\\ \hline \color{red}1 & \color{red}-\\ 2 & 1\\ 3 & 2\\ 4 & 3\\ 5 & 4\\ \color{red}- & \color{red}5 \end{array}\\ \hline \text{Total $:4$ possibilities} \end{array}$
Total possible ways of picking up the cards $= 5 \times 4 = 20$
Thus, the required Probability $= \dfrac{\text{favorable ways}}{\text{total possible ways}}= \dfrac{4}{20} = \dfrac 15$
Option A is correct
How to solve the question if it would be only HIGHER and not ONE HIGHER?
Possible combinations :-
$P(First\ card\ is\ HIGHER\ than\ second\ one\ )=\left [ \left ( \frac{1}{5} \times \frac{4}{4}\right ) + \left ( \frac{1}{5} \times \frac{3}{4}\right ) + \left ( \frac{1}{5} \times \frac{2}{4}\right ) + \left ( \frac{1}{5} \times \frac{1}{4}\right ) \right]\\ =\frac{1}{2}$
@amarVashishth
I have a basic doubt. It is just given that we have 5 cards from 1-5, so should we assume that they are from same suite? Since nothing mentioned about suites of card, we can also have a sample space as selecting card with no.1 from either of the suite thus making sample space as 4*4*4*4*4?
Thanks