orrect answerA 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?

Step 1: Identify the total number of balls. The bag contains 2 red, 3 green, and 2 blue balls. So, the total number of balls is 2 + 3 + 2 = 7 balls.

Step 2: Identify the total number of ways to draw 2 balls from 7. This is a combination problem, because the order in which we draw the balls does not matter. The formula for combinations is C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial.

So, the total number of ways to draw 2 balls from 7 is C(7, 2) = 7! / [2!(7-2)!] = 21 ways.

Step 3: Identify the total number of ways to draw 2 balls that are not blue. There are 5 balls that are not blue (2 red and 3 green). So, the total number of ways to draw 2 balls that are not blue is C(5, 2) = 5! / [2!(5-2)!] = 10 ways.

Step 4: Calculate the probability. The probability that none of the balls drawn is blue is the number of ways to draw 2 balls that are not blue divided by the total number of ways to draw 2 balls.

So, the probability is 10 / 21 = 0.4762, or approximately 47.62% when expressed as a percentage.