Solution
= Solution
The total number of two-fish hauls is $\binom{10}{2}=45$. Selecting both from the eight carp gives
$$
\boxed{\mathbb P(X=0)=\frac{\binom20\binom82}{45}=\frac{28}{45}.}
$$
= Solution
The total number of two-fish hauls is $\binom{10}{2}=45$. Selecting both from the eight carp gives
$$
\boxed{\mathbb P(X=0)=\frac{\binom20\binom82}{45}=\frac{28}{45}.}
$$