Two cards are drawn' successively with replacement from a well shuffled pack of 52 cards. Then mean of number of kings is

Let X denote the number of king.
Since the card is drawn twice, 0,1 and 2 are possible values of X .
Probability of getting a king in a single draw of \(a\) card is \(p=\frac{4}{52}=\frac{1}{13}, q=\frac{12}{13}\)
\(\begin{aligned}
& P[X=0]=P[\text { no king }]=q q=q^2=\frac{144}{169} \\
& \mathrm{P}[\mathrm{X}=1]=\mathrm{P}[\text { one king }]=\mathrm{pq}+\mathrm{qp}=2 \mathrm{pq}=\frac{24}{169} \\
& \mathrm{P}[\mathrm{X}=2]=\mathrm{P}[\text { two kings }]=\mathrm{pp}=\mathrm{p}^2=\frac{1}{169} \\
& \text { Mean }=\sum x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} \\
& =0 \times \frac{144}{169}+1 \times \frac{24}{169}+2 \times \frac{1}{169} \\
& =\frac{26}{169}=\frac{2}{13}
\end{aligned}\)
Alternate method:
Let getting a king be the success.
\(p=\frac{1}{13}, q=\frac{12}{13}\)
Since two cards are drawn with replacement.
\(\therefore \quad n=2\)
Here, r.v. \(X \sim B\left(\frac{1}{13}, 2\right)\)
\(\therefore \quad \text { mean }=\mathrm{np}=\frac{2}{13}\)