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 queens is
$\frac{1}{13}$
$\frac{1}{169}$
$\frac{2}{13}$
$\frac{4}{169}$
Solution
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
$\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\sum x \cdot \mathrm{P}(x) \\
& =0 \times \frac{144}{169}+1 \times \frac{24}{169}+2 \times \frac{1}{169} \\
& =\frac{26}{169} \\
\mathrm{E}(\mathrm{X}) & =\frac{2}{13}
\end{aligned}$