Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Then mean of number…
Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Then mean of number of tens is
$\frac{1}{13}$
$\frac{1}{169}$
$\frac{2}{13}$
$\frac{4}{169}$
Solution
Probability of getting ten $=\frac{4}{52}=\frac{1}{13}$
$\therefore \quad$ Probability of getting a card without ten $=\frac{12}{13}$
Let random variable $\mathrm{X}$ denotes the number of tens.
$\therefore \quad$ 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}
$\therefore \quad$ Required mean
$\begin{aligned}
& =0+1 \times\left(\frac{12}{13 \times 13}+\frac{12}{13 \times 13}\right)+2 \times\left(\frac{1}{13} \times \frac{1}{13}\right) \\
& =\frac{24}{169}+\frac{2}{169} \\
& =\frac{26}{169} \\
& =\frac{2}{13}
\end{aligned}$