Probability Distribution sum The sum regarding probability distribution is:

There is a group of 50 people who are patriotic out of which 20 believe in non-violence. Two persons are selected at random out of them. Write the probability distribution for the selected persons who are non-violent. Also find the mean of the distribution.

My solution:
$$\begin{array}{r|lll}
x & 0 & 1 & 2 \\
P(X) & {n \choose 0}\frac25^0\frac35^2 & {n \choose 1}\frac25^1\frac35^1 & {n \choose 2}\frac25^2\frac35^0 \\
P(X) & \frac9{25} & \frac{12}{25} & \frac4{25}
\end{array}$$
$$\text{Mean}=\dfrac{9}{25} + 2 \cdot \dfrac{4}{25} = \dfrac{17}{25}$$
However, my book is suggesting $\dfrac{196}{245}$.
Which one is correct?
 A: The default interpretation, I think, is that the people are selected without replacement. Then for example $\Pr(X=0)=\frac{30}{50}\cdot \frac{29}{49}$. Similarly, $\Pr(X=1)=2\cdot \frac{30}{50}\cdot \frac{20}{49}$ and $\Pr(X=2)=\frac{20}{50}\cdot \frac{19}{49}$.
Now one can find the mean in the usual way.  You should get (after simplification) $\frac{4}{5}$.
Remark: Your calculation of the mean, even under the assumption of sampling with replacement, had a little slip. You should have obtained $1\cdot\frac{12}{25}+2\cdot \frac{4}{25}$.  Note that this simplifies to $\frac{4}{5}$.  Interestingly, this is exactly the same as the result we obtained earlier. 
That is not an accident.  Suppose that we sample $n$ times, (i) with replacement and (ii) without replacement from a population of $N$ objects, $g$ of which are good. Then in Case (i), the number of good in the sample has binomial distribution. In Case (ii), the number of good in the sample has hypergeometric distribution.
However, in both cases, the mean number of good is $\frac{ng}{N}$.
A: The distribution that should be used is the Hypergeomteric distribution, which is given by the formula 
$$P(X=k)=\frac{{r \choose k}{ N-r \choose n-k}}{N \choose n}$$
where (in your case) $N=50$ is the total population, $k$ is the number of non-violent persons in your selection $k\in \{0,1,2\}$, $r=20$ is the number of non-violent people in the population and $n=2$ is the number of people drawn without replacement.
The mean will be
$$\mathbb{E}(X)=\sum_{k=0}^2k\cdot P(X=k)=0\cdot\frac{{20 \choose 0}{ 30 \choose 2}}{50 \choose 2}+1\cdot\frac{{20 \choose 1}{ 30 \choose 2}}{50 \choose 1}+2\cdot\frac{{20 \choose 2}{ 30 \choose 0}}{50 \choose 2}=0.8=\frac{196}{245}$$
A: Among the $\binom{50}{2}$ possible couples we can select, $\binom{20}{2}$ of them are made of two non-violent people, $\binom{30}{2}$ of them are made of violent people and $\binom{50}{2}-\binom{30}{2}-\binom{20}{2}=20\cdot 30$ of them are mixed, so:
$$\mathbb{P}[X=0]=\frac{\binom{30}{2}}{\binom{50}{2}}=\frac{87}{245},$$ 
$$\quad \mathbb{P}[X=1]=\frac{20\cdot 30}{\binom{50}{2}}=\frac{120}{245},$$
$$ \mathbb{P}[X=2]=\frac{\binom{20}{2}}{\binom{50}{2}}=\frac{38}{245}$$
and:
$$\mathbb{E}[X]=\frac{87\cdot 0+120\cdot 1+38\cdot 2}{245}=\color{red}{\frac{196}{245}}=\frac{4}{5}.$$
