Probability of caugh at least 1 of one type of fish In the lake we have got 3 types of fish:
k - number of roach
2k - number of crucian
4k - number of perch
Mr Smith caught 7 fish. What is a probability that Mr Smith caught at least 1 roach.
My solution:
$P(\Omega)= {{7k} \choose {7}}$
$P(A) = \frac { {{k} \choose {1}}}{{{7k} \choose {7}}}$
Is it ok?
 A: The total number of combinations: $\dbinom{7k}{7}$

The total number of combinations without roach fish: $\dbinom{6k}{7}$

The probability for not catching any roach fish: $\dfrac{\dbinom{6k}{7}}{\dbinom{7k}{7}}$

The probability for catching at least one roach fish: $1-\dfrac{\dbinom{6k}{7}}{\dbinom{7k}{7}}$

You can further simplify the above result to: $1-\dfrac{(6k)!(7k-7)!}{(7k)!(6k-7)!}$
A: If $k$ is large, (and as usual we assume independence) we have the simpler approximate answer $1-\left(\frac{6}{7}\right)^7$.
We have probability exactly  $1-\left(\frac{6}{7}\right)^7$ if Mr. Smith catches and releases, a real world example of sampling with replacement! That assumes that fish when caught and released don't learn from their experience. 
A: The probability of not catching a roach in $7$ tries can be computed as follows: 
$$
\begin{align}
\overbrace{\frac{6k}{7k}}^\text{first fish}\ \overbrace{\frac{6k-1}{7k-1}}^\text{second fish}\cdots\overbrace{\frac{6k-6}{7k-6}}^\text{seventh fish}
&=\frac{(6k)!}{(6k-7)!}\frac{(7k-7)!}{(7k)!}\\
&=\frac{\binom{6k}{7}}{\binom{7k}{7}}
\end{align}
$$
For the first fish caught, there are $6k$ other fish out a total of $7k$. After not catching a roach on the first try, the second try there are $6k-1$ other fish out of a total of $7k-1$. After not catching a roach on the second try, the third try there are $6k-2$ other fish out of a total of $7k-2$. Et cetera.
Thus, the probability of catching at least $1$ roach is the complement of not catching one:
$$
1-\frac{\binom{6k}{7}}{\binom{7k}{7}}
$$
