Question

From a pack of $52$ cards, two are drawn with replacement.

The probability, that the first is a diamond and the second is a king is:

• A. $\frac{1}{26}$
• B. $\frac{17}{2704}$
• C. $\frac{1}{52}$
• D. None of these.

Answer 1

The correct option is C

$\frac{1}{52}$

The explanation for the correct option:

Step: 1 Find the probability of getting a diamond at the first attempt:

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

There are $13$ diamonds that pack.

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

Event of choosing a diamond card $n\left(D\right)={}^{13}C_1$

If one card is drawn at random, then the probability of getting a diamond is $P\left(D\right)$

$\Rightarrow$$\frac{n\left(D\right)}{n\left(S\right)}=\frac{{}^{13}C_1}{{}^{52}C_1}$ $\left[\mathbf{\because }\mathit{P}\mathbf{\left(}\mathit{D}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{\left(}\mathbf{D}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}\right]$

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

Step: 2 Find the probability of getting a king in the second attempt:

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

There are $4$ kings in that pack.

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

Event of choosing a king card $n\left(K\right)={}^{4}C_1$

If one card is drawn at random, then the probability of getting a king is $P\left(K\right)$

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

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

Step: 3 Find the required probability:

Therefore, the probability of getting the first as a diamond and the second as a king is

$\begin{array}{rcl}P\left(K\right)\times P\left(D\right)& =& \frac{1}{13}\times \frac{1}{4}\\ & =& \frac{\mathbf{1}}{\mathbf{52}}\end{array}$

Hence, Option(C) is the correct answer.