Question

" A bag contains 4 red, 3 white and 2 blue balls. 3 balls are drawn randomly. Determine the number of ways of selecting at least 1 white marble. "

Answer 1

First consider the probability of drawing a red ball at the first draw, a black at the second and a blue at the third. The first probability is 2/9, the second, conditional on the first is 3/8 and the third conditional on the other two is 4/7. The product is (2*3*4)/(9*8*7). It is easy to see that any other order is equally likely (the only difference is the order of the factors in the numerator). So adding 6 equally likely cases we obtain (6*2*3*4)/(9*8*7) = 2/7.

Combinations approach-

Let us assume that all balls are unique. 
There are a total of 9 balls.

Total ways = 3 balls can be chosen in 9C3 ways = 9(6!3!) = 9*8*7/3*2*1 = 84
Favorable ways = 1 Red ball, 1 Black ball, and 1 Blue Ball = 2*3*4 = 24