Two cards are drawn successively with replacement from a well shuffled pack of 52 cards, then mean of number of queens is

Total number of cards \(=52\)
Total number of queens \(=4\)
Probability of getting a queen
\(P(\text { queen })=\frac{4}{52}=\frac{1}{13}\)
Probability of not getting a queen
\(\mathrm{P}(\text { non queen })=\frac{48}{52}=\frac{12}{13}\)
Let \(\mathrm{X}\) be a random variable such that \(\mathrm{X}=\) number of queens in 2 draws
Case I: No queens are drawn \((X=0)\) \(\mathrm{P}(\mathrm{X}=0)=\mathrm{P}(\) non queen \() \times \mathrm{P}(\) non queen \()\)
\(=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}\)
Case II: One queen is drawn \((\mathrm{X}=1)\) \(\mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\) non queen and queen \()\) or \(\mathrm{P}(\) queen and non queen)
\(\begin{aligned}
& =\frac{12}{13} \times \frac{1}{13}+\frac{1}{13} \times \frac{12}{13} \\
& =\frac{24}{169}
\end{aligned}\)
Case III: Two queens are drawn \((\mathrm{X}=2)\)
\(\begin{aligned}
\mathrm{P}(\mathrm{X}=2) & =\mathrm{P}(\text { queen }) \times \mathrm{P}(\text { queen }) \\
& =\frac{1}{13} \times \frac{1}{13} \\
& =\frac{1}{169}
\end{aligned}\)
Required Mean is
\(\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\sum x \cdot \mathrm{P}(x) \\
& =0 \times \frac{144}{169}+1 \times \frac{24}{169}+2 \times \frac{1}{169} \\
& =\frac{26}{169} \\
\mathrm{E}(\mathrm{X}) & =\frac{2}{13}
\end{aligned}\)