$\dfrac{\_4C\_2 \times~~\_3C\_1}{\_{10}C\_3}=\dfrac{6\times3}{120}=\dfrac{3}{20}$. 
In a box, there are $10$ balls of which $4$ are red, $3$ are black and $3$ are white. What is the probability of picking up $3$ balls randomly such that there are $2$ red balls and $1$ balck ball? 
a. $\frac12$ 
b. $\frac13$ 
c. $\frac1{12}$ 
d. $\dfrac{3}{20}$ 