<s>[INST] A Basic Probability Question - I am getting the wrong answer Problem:
In a deck of $52$ cards there are $4$ kings. A card is drawn at random
from the deck and its face value noted; then the card is returned. This
procedure is followed $4$ times. Compute the probability that there are
exactly $2$ kings in the $4$ selected cards if it is known that there is at
least one king in those selected.
Answer:
Let $p$ be the probability we seek. Let $p_2$ be the probability that we
draw exactly $2$ kings. Let $p_1$ be the probability that we draw at least
$1$ king.
\begin{eqnarray*}
p &=& \frac{p_2}{p_1} \\
p_2 &=& {13 \choose 2}{(\frac{4}{52})^2}{(\frac{48}{52})}^2 \\
{13 \choose 2} &=& \frac{13(12)}{2(1)} = 13(6) \\
p_2 &=& 13(6){(\frac{1}{13})^2}{(\frac{12}{13})}^2 = \frac{6(12)^2}{13^3} \\
\end{eqnarray*}
Let $p_0$ be the probability that we draw no kings.
\begin{eqnarray*}
p_0 &=& (\frac{48}{52})^4 = (\frac{12}{13})^4 \\
p_1 &=& 1 - p_0 = 1 - (\frac{12}{13})^4 = \frac{13^4 - 12^4}{13^4} \\
p &=& \frac{ \frac{6(12)^2}{13^3} }{ \frac{13^4 - 12^4}{13^4} } =
\frac{6(13)(12)^2}{13^4 - 12^4} \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
p = \frac{6(12)^2}{13^4 - 12^4} \\
\end{eqnarray*}
I am hoping that somebody can tell me where I went wrong.
~
 [/INST] $p_2$ the probability that exactly two of four draws will be kings (one of thirteen cards in a suit) is:$$p_2= \binom{\color{red}{4}}{2}\left(\frac{1}{13}\right)^2\left(\frac{12}{13}\right)^2$$
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