Two cards are drawn with replacement from a well shuffled deck of $52$ cards. Find the mean and variance for the number of aces.
Let $X$ be a random variable which represents drawing an ace from a well shuffled deck of $52$ cards.
Then $X$ takes values $0, \; 1, \; 2$.
$2$ cards can be selected (with replacement) from a pack of $52$ cards in $52 \times 52$ ways.
Now, $P \left(X = 0\right) = $ P(No ace is selected)
The first card can be selected in $48$ ways; the second card can be selected in $48$ ways.
$\therefore \;$ $P \left(X = 0\right) = \dfrac{48 \times 48}{52 \times 52} = \dfrac{12 \times 12}{13 \times 13} = \dfrac{144}{169}$
$P \left(X = 1\right) = $ P(only one of the two cards is an ace)
$1$ ace can be selected from $4$ aces in $4$ ways; $1$ card (other than ace) can be selected in $48$ ways.
Amongst themselves, the two cards can be selected in $2$ ways.
$\therefore \;$ $P \left(X = 1\right) = 2 \times \dfrac{4 \times 48}{52 \times 52} = 2 \times \dfrac{1}{13} \times \dfrac{12}{13} = \dfrac{24}{169}$
$P \left(X = 2\right) = $ P(both cards are ace cards)
The first card can be selected in $4$ ways; the second card can be selected in $4$ ways.
$\therefore \;$ $P \left(X = 2\right) = \dfrac{4 \times 4}{52 \times 52} = \dfrac{1}{13} \times \dfrac{1}{13} = \dfrac{1}{169}$
$\therefore \;$ We have
| $X$ | $0$ | $1$ | $2$ |
|---|---|---|---|
| $P \left(X = x\right)$ | $\dfrac{144}{169}$ | $\dfrac{24}{169}$ | $\dfrac{1}{169}$ |
Mean value of $X = E \left(X\right) = \sum \limits_{i} x_i p_i$
$\therefore \;$ $E \left(X\right) = \left(0 \times \dfrac{144}{169}\right) + \left(1 \times \dfrac{24}{169}\right) + \left(2 \times \dfrac{1}{169}\right) = \dfrac{26}{169} = \dfrac{2}{13}$
Variance of $X = E \left(X^2\right) - \left[E \left(X\right)\right]^2$
Now, $E \left(X^2\right) = \sum \limits_{i} p_i x_{i}^{2}$
$\therefore \;$ $E \left(X^2\right) = \left(0^2 \times \dfrac{144}{169}\right) + \left(1^2 \times \dfrac{24}{169}\right) + \left(2^2 \times \dfrac{1}{169}\right) = \dfrac{28}{169}$
$\left[E \left(X\right)\right]^2 = \left(\dfrac{2}{13}\right)^2 = \dfrac{4}{169}$
$\therefore \;$ Variance of $X = \dfrac{28}{169} - \dfrac{4}{169} = \dfrac{24}{169}$