Question

Two cards are drawn without replacement from a well shuffled pack. Find the probability that one of them is an ace of heart

• A. $\frac{1}{25}$
• B. $\frac{1}{26}$
• C. $\frac{1}{52}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{26}$

Find the required probability:

There are $52$ cards in a pack of cards, but there is only $1$ ace of hearts.

Let $\mathrm{A}$ be the event of getting an ace of hearts

$\Rightarrow$ $\mathrm{p}\left(\mathrm{A}\right)=\frac{1}{52}$

Let $\mathrm{B}$ be the event of not getting an ace of hearts

$\Rightarrow$ $\mathrm{p}\left(\mathrm{B}\right)=\frac{51}{52}$

Two card are drawn. There are two ways of drawing the cards such that at least one of them is an ace of hearts

Case $1$: Let ${\mathrm{C}}_1$ be the event where we draw the first card an ace of hearts and the second card as non ace of hearts

$\mathrm{p}\left({\mathrm{C}}_1\right)=\frac{1}{52}\times \frac{51}{51}$ [$51$ non ace of hearts remain in the remaining $51$ cards]

$\Rightarrow$ $\mathrm{p}\left({\mathrm{C}}_1\right)=\frac{1}{52}$

Case $2$: Let ${\mathrm{C}}_2$ be the event where we draw the first card as non ace of hearts and the second card as ace of hearts

$\mathrm{p}\left({\mathrm{C}}_2\right)=\frac{51}{52}\times \frac{1}{51}$ [$1$ ace of hearts remain in the remaining $51$ cards]

$\Rightarrow$ $\mathrm{p}\left({\mathrm{C}}_2\right)=\frac{1}{52}$

Therefore, the total probability is $\mathrm{p}\left({\mathrm{C}}_1\right)+\mathrm{p}\left({\mathrm{C}}_2\right)=\frac{1}{52}+\frac{1}{52}=\frac{1}{26}$

So, the probability of getting an ace of hearts on drawing two cards from a full pack of cards is $\frac{1}{26}$.

Hence, option (B) is the correct answer.