A deck of $5$ cards (each carrying a distinct number from $1$ to $5$) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card?
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