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:
$$\begin{align}
\mathsf P(N_R=N_r) & =\frac{1}{10}\frac{6!\,(10-1)!}{(6-1)!\,10!}+\frac 5{10}\frac{6!\,(10-2)!}{(6-2)!\,10!}+\frac{1}{5}\frac{6!\,(10-3)!}{(6-3)!\,10!}+...
\\[1ex] & 
\end{align}$$

Let $A = \{\text{Choose only red}\}.$ Let $X$ be the number of balls drawn, and let $R$ be the number of red balls drawn. Then
\begin{align*}
P(A) &=\sum_{k = 0}^{6} P(R  = k|X = k)P(X = k)\tag 1\\
&=\sum_{k = 0}^{6} \frac{\binom{6}{k}}{\binom{10}{k}}p_k\\
&=\frac{\binom{6}{1}}{\binom{10}{1}}(.1)+\frac{\binom{6}{2}}{\binom{10}{2}}(.5)+\frac{\binom{6}{3}}{\binom{10}{3}}(.2)+\frac{\binom{6}{4}}{\binom{10}{4}}(.1)+\frac{\binom{6}{5}}{\binom{10}{5}}(0)+\frac{\binom{6}{6}}{\binom{10}{6}}(0)\tag2\\
&= \frac{187}{700},
\end{align*}
where in $(1)$ it has to be the case that the number the random number you draw has to equal the number of red balls you draw; in $(2)$, $p_5 = p_6 = 0$ since according to the distribution of $X$, you cannot draw 5 or 6 times.
So, our answers agree, almost. If you complete your calculation (since you didn't post it), our answers should agree.
