A box contains 2 red, 3 green and 2 blue balls. If two balls are drawn at random what is the probability that none of the balls drawn is blue?

Total number of balls =2+3+2=7 balls
Number of ways of drawing 2 balls = ${}^{7}c_2$=21

∵ None of the balls drawn is blue.

So, number of ways of drawing 2 balls from 2 red and 3 green balls   = ${}^5{c}_2=10$

Hence, required probability =$\frac{10}{21}$