In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red?

Total balls = 6 red + 9 green = 15. P(at least one red) = 1 - P(both green). P(both green) = (9/15) × (8/14) = 72/210 = 12/35. Therefore, P(at least one red) = 1 - 12/35 = 23/35. Alternatively, we can calculate directly: cases are (1 red, 1 green) or (2 red). P(1 red, 1 green) = (6/15)×(9/14)×2 = 108/210 = 18/35. P(2 red) = (6/15)×(5/14) = 30/210 = 5/35. Total = 18/35 + 5/35 = 23/35.