Question

Two cards are drawn successively with replacement from a well-shuffled deck of $52$ cards, then the mean of the number of aces is

• A. $\frac{1}{13}$
• B. $\frac{3}{13}$
• C. $\frac{2}{13}$
• D. None of these

Answer 1

The correct option is C

$\frac{2}{13}$

Find the required probability:

Step 1: Find the probability of drawing an ace card

Total number of cards $=52$

Total number of aces $=4$

Probability of getting an ace

$\Rightarrow$ $\mathrm{p}\left(\mathrm{ace}\right)=\frac{4}{52}=\frac{1}{13}$

Probability of not getting an ace

$\Rightarrow$ $\mathrm{p}(\mathrm{non}\mathrm{ace})=\frac{48}{52}=\frac{12}{13}$

Step 2: Find the mean value of the distribution.

Let $\mathrm{X}$ be a random variable such that,

$\mathrm{X}=$number of aces in $2$ draws

Case $1$: No aces are drawn $\left(\mathrm{X}=0\right)$

$\mathrm{p}\left(\mathrm{X}=0\right)=\mathrm{p}\left(\mathrm{non}\mathrm{ace}\right)\times \mathrm{p}\left(\mathrm{non}\mathrm{ace}\right)$

$=\frac{12}{13}\times \frac{12}{13}$

$\Rightarrow$ $\mathrm{p}\left(\mathrm{X}=0\right)=\frac{144}{169}$

Case $2$: One ace is drawn $\left(\mathrm{X}=1\right)$

$\text{p}\left(\mathrm{X}=1\right)=\mathrm{p}\left(\left(\mathrm{non}\mathrm{ace}\mathrm{and}\mathrm{ace}\right)\mathrm{or}\left(\mathrm{ace}\mathrm{and}\mathrm{non}\mathrm{ace}\right)\right)$

$=\frac{12}{13}\times \frac{1}{13}+\frac{1}{13}\times \frac{12}{13}$

$\Rightarrow$ $\text{p}\left(\mathrm{X}=1\right)=\frac{24}{169}$

Case $3$: Two aces are drawn $\left(\mathrm{X}=2\right)$

$\mathrm{p}\left(\mathrm{X}=2\right)=\mathrm{p}\left(\mathrm{ace}\right)\times \mathrm{p}\left(\mathrm{ace}\right)$

$=\frac{1}{13}\times \frac{1}{13}$

$\Rightarrow$ $\mathrm{p}\left(\mathrm{X}=2\right)=\frac{1}{169}$

The mean or the expected value of such probability distribution is given as

$\mathrm{E}\left(\mathrm{X}\right)=\sum _{}\mathrm{X}.\mathrm{p}\left(\mathrm{X}\right)$

$\Rightarrow$ $\mathrm{E}\left(\mathrm{X}\right)=0\times \frac{144}{169}+1\times \frac{24}{169}+2\times \frac{1}{169}$

$\Rightarrow$ $\mathrm{E}\left(\mathrm{X}\right)=\frac{26}{169}$

$\Rightarrow$ $\mathrm{E}\left(\mathrm{X}\right)=\frac{2}{13}$

Hence, the mean number of aces drawn is $\frac{2}{13}$.

Hence, option (C) is the correct answer.