The total number of ways of selecting $4$ elk from $20$ elk is ${}^{20}C_4$.
Out of the $20$ elk, $5$ are tagged and $15$ are untagged.
We want exactly $2$ of the $4$ captured elk to be tagged.
The number of ways to choose $2$ tagged elk from $5$ tagged elk is ${{}^5C_2}$.
The number of ways to choose the remaining $2$ elk from the $15$ untagged elk is ${{}^{15}C_2}$.
So, the number of favorable outcomes is
${}^5C_2 \cdot {}^{15}C_2$.
Therefore, the required probability is
$P(\text{exactly }2\text{ tagged})=\dfrac{{}^5C_2\cdot{}^{15}C_2}{{}^{20}C_4}$.
Now,
${}^5C_2=10$, $\quad$ ${}^{15}C_2=105$, $\quad$ ${}^{20}C_4=4845$.
Hence,
$P(\text{exactly }2\text{ tagged})=\dfrac{10\cdot105}{4845} =\dfrac{1050}{4845} =\dfrac{70}{323}$.
Thus, the probability that exactly $2$ of the $4$ captured elk have been tagged is $\dfrac{70}{323}$.
