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

Probability of getting ten \(=\frac{4}{52}=\frac{1}{13}\)
\(\therefore\) Probability of getting a card without ten \(=\frac{12}{13}\)
Let random variable \(\mathrm{X}\) denotes the number of tens.
\(\therefore\) Possible values of \(\mathrm{X}\) are \(0,1,2\)
Consider following probability distribution table.
\(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 \\
\hline \mathrm{P}(\mathrm{X}=x) & \frac{12}{13} \times \frac{12}{13} & \frac{1}{13} \times \frac{12}{13}+\frac{12}{13} \times \frac{1}{13} & \frac{1}{13} \times \frac{1}{13} \\
\hline
\end{array}\)
\(\therefore\) Required mean
\(=0+1 \times\left(\frac{12}{13 \times 13}+\frac{12}{13 \times 13}\right)+2 \times\left(\frac{1}{13} \times \frac{1}{13}\right) \)
\( =\frac{24}{169}+\frac{2}{169} \)
\( =\frac{26}{169} \)
\( =\frac{2}{13}\)