This problem is an example of hypergeometric distribution..It is similar to binomial distribution but differs with respect to following things :
a) The population being considered here is finite and the mode of event is "without replacement".
b) The probability changes for each individual event , which violates the assumption of Binomial distribution that probabilty should remain the same.
So here in total we have 10 balls , which is a finite population and out of which we have 4 red and 6 black balls..Now according to the question , favorable event is the one in which we have one red balls and two black balls..
Total number of events = Number of ways in which we can pick 3 out of 10 balls = 10C3
Now number of favorable events = Number of ways 1 red ball can be picked out of 4 red balls * Number of ways 2 black balls can be picked from 6 balls = 4C1 * 6C2
Hence required probability = ( 4C1 * 6C2 ) / 10C3
= ( 4 * 15 * 6 ) / ( 10 * 9 * 8 )
= 1 / 2
Ref : https://en.wikipedia.org/wiki/Hypergeometric_distribution