Question

From a pack of $52$ cards, two cards are drawn in succession one by one without replacement. The probability that both are aces is

• A. $\frac{2}{13}$
• B. $\frac{1}{51}$
• C. $\frac{1}{221}$
• D. $\frac{2}{21}$

Answer 1

The correct option is C

$\frac{1}{221}$

The explanation for the correct option:

Step: 1 Find the probability of getting an ace at the first attempt:

There are $52$ cards in a pack of cards.

There are $4$ aces in that pack.

Event of choosing one card $n\left(S\right)={}^{52}C_1$

Event of choosing an ace $n\left(A_1\right)={}^{4}C_1$

If one card is drawn at random, then the probability of getting an ace is $P\left(A_1\right)$

$\Rightarrow$$\frac{n\left(A_1\right)}{n\left(S\right)}=\frac{{}^{4}C_1}{{}^{52}C_1}$ $\left[\mathbf{\because }\mathit{P}\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{\left(}\mathbf{A}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}\right]$

$$\begin{gathered} =\frac{4}{52} \\ =\frac{1}{13} \end{gathered}$$

Step: 2 Find the probability of getting a second ace without replacement:

Now, there are $51$ cards in that pack.

There are $3$ aces in that pack.

Event of choosing one card $n\left(S\right)={}^{51}C_1$

Event of choosing a second ace $n\left(A_2\right)={}^{3}C_1$

If one card is drawn at random, then the probability of getting a second ace without any replacement is $P\left(A_2\right)$

$\Rightarrow$$\frac{n\left(A_2\right)}{n\left(S\right)}=\frac{{}^{3}C_1}{{}^{51}C_1}$ ($\because$one card is drawn at random)

$$\begin{gathered} =\frac{3}{51} \\ =\frac{1}{17} \end{gathered}$$

Step: 3 Find the required probability:

Therefore, the probability of getting two aces without replacement is equal to multiplying of the probability of getting an ace and the probability of getting a second ace without any replacement

$\begin{array}{rcl}P\left(A_1\right)\times P\left(A_2\right)& =& \frac{1}{13}\times \frac{1}{17}\\ & =& \frac{\mathbf{1}}{\mathbf{221}}\end{array}$

Hence, Option(C) is the correct answer.