Question 306204
Let nCk = number of ways to pick up k items from a set of n items.
Of course you should already know that {{{nCk = n!/(k!*(n-k)!)}}} (*)
Bag consists of 2 red balls (R), 3 blue balls (B) and 5 green balls (G)
How many ways to pick up any 3 balls from the bag: 10C3 
a> three drawn balls are of different colors --> this means to pick 1R, 1B and 1G
--> there are totally of 2C1*3C1*5C1 --> probability of 2C1*3C1*5C1/10C3
b> two of three drawn balls are of the same color.
There are 3 cases: the two balls of same color are red(1), blue (2) and green(3).
(1) means 2R and any one out of the 8 other balls (3B and 5G) --> 2C2*8C1
(2) means 2B and any one out of the 7 other balls (2R and 5G) --> 3C2*7C1
(3) means 2G and any one out of the 5 other balls (2R and 3B) --> 5C2*5C1
Thus, there are totally 2C2*8C1 + 3C2*7C1 + 5C2*5C1 ways to pick out 3 balls, 2 of which are of the same color --> the probability of (2C2*8C1 + 3C2*7C1 + 5C2*5C1)/10C3
c> all the three balls are of the same color.
This one is a bit more interesting. Obviously, the three drawn balls can't be red since there are only 2 red balls. Thus, there are only 2 cases: all three balls of the same color are blue(4) and green(5)
(4) means to pick 3B out of the 3 blue balls --> 3C3
(5) means to pick 3G out of the 5 green balls --> 5C3
Thus, there are totally 3C3 + 5C3 ways to pick out 3 balls, 2 of which are of the same color --> the probability of (3C3 + 5C3)/10C3
You surely can use the (*) for the final calculations, right?