ProbabilityA 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?Options10/215/211/2111/21

To solve this problem, we need to understand that the probability of an event is the number of ways that event can occur divided by the total number of outcomes.

Step 1: Calculate the total number of balls in the bag. There are 2 red balls, 3 green balls, and 2 blue balls. So, the total number of balls is 2 + 3 + 2 = 7 balls.

Step 2: Calculate 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: Calculate the number of ways to draw 2 balls that are not blue. Since there are 2 blue balls, there are 7 - 2 = 5 balls that are not blue. So, the 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 that none of the balls drawn is blue. The probability 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.

So, the correct answer is 10/21.