A bag contains seven green, six blue and five yellow balls. Two balls are drawn randomly. What is the probability that neither of the balls drawn is a green ball?

Sure, here is the step by step solution:

Step 1: Calculate the total number of balls in the bag. The total number of balls is 7 (green) + 6 (blue) + 5 (yellow) = 18 balls.

Step 2: Calculate the total number of ways to draw 2 balls out of 18. This can be calculated using the combination formula C(n, r) = n! / [(r!(n-r)!)], where n is the total number of items, and r is the number of items to choose. So, the total number of ways to draw 2 balls out of 18 is C(18, 2) = 18! / [(2!(18-2)!)] = 153.

Step 3: Calculate the total number of ways to draw 2 balls that are not green. This means we are only considering the blue and yellow balls, which total 6 (blue) + 5 (yellow) = 11 balls. So, the total number of ways to draw 2 balls out of 11 is C(11, 2) = 11! / [(2!(11-2)!)] = 55.

Step 4: Calculate the probability that neither of the balls drawn is a green ball. This is the number of ways to draw 2 balls that are not green divided by the total number of ways to draw 2 balls. So, the probability is 55 / 153 = 0.359 (rounded to three decimal places).

So, the probability that neither of the balls drawn is a green ball is approximately 0.359, or 35.9%.