Find prob. that only select red balls from $n$ (red+blue) balls There are 4 blue balls and 6 red balls(total 10 balls). $X$ is a random variable of the number of selected balls(without replacement), in which
$$P(X=1)=0.1$$
$$P(X=2)=0.5$$
$$P(X=3)=0.2$$
$$P(X=4)=0.1$$
$$P(X=10)=0.1$$
Then, what is probability of only selecting red balls?
This is what I have tried:
The (conditional) probability that all $r$ of the balls are selected from the red is just: ${6\choose r}\big/{10\choose r}$, for $0\leq r\leq 6$ , and $0$ elsewhere. 
That is, let $N_R$ be the number of red balls selected, and $N_r$ the total number of balls selected, then:
$$\mathsf P(N_r=N_R\mid N_r=r) = \frac{6!/(6-r)!}{10!/(10-r)!} \mathbf 1_{r\in\{1\ldots 6\}}$$
As the number of balls selected is a random variable with the specified distribution, then the probability that all balls selected are red is:
$$\begin{align}
\mathsf P(N_R=N_r) & =\frac{1}{10}\frac{6!\,(10-1)!}{(6-1)!\,10!}+\frac 5{10}\frac{6!\,(10-2)!}{(6-2)!\,10!}+\frac{1}{5}\frac{6!\,(10-3)!}{(6-3)!\,10!}+...
\\[1ex] & 
\end{align}$$

Since there is $6$ red balls, if $10$ balls are selected, probability of selecting only red balls is $0$ and we only have to consider selecting $1,2,3,4$ balls.
Let $R$ be number of red balls selected.
$$\begin{align}P(R=i|X=i)&=\frac{_6P_i}{_{10}P_i}\\
P(R=X)&=\sum\limits_{i=1}^4P(X=i)\cdot P(R=i|X=i)\\
&=0.1\cdot \frac 6{10}+0.5\cdot \frac {6\cdot5}{10\cdot9}+0.2\cdot \frac {6\cdot5\cdot4}{10\cdot9\cdot8}+0.1\cdot \frac {6\cdot5\cdot4\cdot3}{10\cdot9\cdot8\cdot7}\\
&=\frac{187}{700}\end{align}$$
