<s>[INST] 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?
 [/INST] 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:
$$\mathbb{E}[X]=\frac{87\cdot 0+120\cdot 1+38\cdot 2}{245}=\color{red}{\frac{196}{245}}=\frac{4}{5}.$$
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