A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue? A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

I am helpless regarding this. I don't know how to solve it. My teacher asked me to solve it by finding the probability that the balls drawn are blue and then subtracting it from 1. But I want to solve it straight forward and directly. Is it possible? If yes, how?
Assumption: The first ball drawn is not put back in the bag.

Let E be the event that none of the two picked balls are blue. Further let $E_1$ and $E_2$ be the events that the first ball picked is not blue and the second ball picked is not blue respectively.

$P(E) = P(E_1 \cap E_2) = P(E_1)P(E_2 | E_1)$

$\implies P(E) = \left( {5 \choose 1}*1/7 \right) \left( {4 \choose 1}*1/6 \right)$

$\implies P(E) =  10/21 $
In a bag there are 4 red balls ,2green balls, and 3yellow balls .if two balls are drawn at random ,what is the probability that at least one of the balls is green in color.