A bag contains 2 red balls, 3 green balls and 2 blue balls. Two balls are drawn at random from the bag without replacement. What is the probability that none of the balls drawn is blue, that is, both balls are either red or green?

Introduction / Context:
This question is a probability problem involving drawing balls from a bag without replacement. The bag contains balls of three different colors, and we are required to find the probability that neither of the two drawn balls is blue. This tests understanding of combinations and probabilities in the context of sampling without replacement, where order does not matter for the subset chosen.

Given Data / Assumptions:

• Red balls in the bag = 2.
• Green balls in the bag = 3.
• Blue balls in the bag = 2.
• Total balls in the bag = 2 + 3 + 2 = 7.
• Two balls are drawn at random without replacement.
• We want the probability that both balls drawn are not blue.

Concept / Approach:
The sample space consists of all possible unordered pairs of balls drawn from the 7 balls. We use combinations to count these pairs. For the event that no drawn ball is blue, both balls must come from the 5 balls that are red or green. Hence we count combinations chosen from these 5 non blue balls. The probability is the ratio of the number of favorable pairs to the total number of pairs from all 7 balls.

Step-by-Step Solution:
Step 1: Compute total number of unordered ways to draw 2 balls from 7 balls: C(7,2) = 21.Step 2: Identify the number of balls that are not blue. There are 2 red and 3 green balls, giving 5 non blue balls in total.Step 3: Compute the number of unordered ways to draw 2 balls from these 5 non blue balls: C(5,2) = 10.Step 4: These 10 combinations represent all favorable outcomes where neither drawn ball is blue.Step 5: The required probability is favorable combinations divided by total combinations, so probability = 10 / 21.Step 6: This fraction is already in simplest form, so the probability that none of the two balls is blue is 10 / 21.
Verification / Alternative check:
We can verify using a complementary approach by finding the probability that at least one ball is blue and then subtracting from 1. The probability that the first ball is blue is 2 / 7, and conditional probabilities for the second draw depend on the first. However, the combination method is more straightforward and avoids order based complications. Checking that C(5,2) plus combinations involving blue balls equals C(7,2) confirms that we have accounted for all possibilities.

Why Other Options Are Wrong:
11/21 would require 11 favorable non blue combinations, which exceeds the 10 that actually exist. 1/2 is a rough guess without considering the exact counts, and 2/7 would correspond to a much smaller set of favorable outcomes. 5/21 is half of the correct numerator and does not match any meaningful selection pattern in this context.

Common Pitfalls:
Students sometimes treat the draws as ordered and then forget to divide out permutations, leading to double counting. Another error is to use simple products of probabilities like (5/7) * (4/6) for non blue draws, which actually gives the same result but may be used incorrectly if the second fraction is miscalculated. Being methodical with combinations ensures that the correct result is obtained even when the numbers are more complex.

Final Answer:
The probability that neither of the two balls drawn is blue is 10/21.