A 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?

Given data

• Red balls = 2
• Green balls = 3
• Blue balls = 2
• Total balls = 2 + 3 + 2 = 7
• Draws = 2 (without replacement)
• Event: none of the drawn balls is blue (i.e., both drawn from only red/green set)

Concept/ApproachUse combinations (sampling without replacement). The favorable outcomes are all 2-ball selections from the non-blue balls. The total outcomes are all 2-ball selections from the full set. Probability = favorable ÷ total.

Step-by-Step calculation Non-blue balls = red + green = 2 + 3 = 5 Favorable ways = C(5, 2) = 5 × 4 ÷ 2 = 10 Total ways = C(7, 2) = 7 × 6 ÷ 2 = 21 P(no blue) = Favorable ÷ Total = 10 ÷ 21

Verification/Alternative Sequential method: First draw non-blue with probability 5÷7; then, without replacement, 4 non-blue remain out of 6 total, so probability 4÷6. Multiply: (5÷7) × (4÷6) = 20÷42 = 10÷21 (order of the two non-blue balls does not matter, and the calculation already accounts for that because draws are successive without replacement). Complement check: P(at least one blue) = 1 − P(no blue) = 1 − 10÷21 = 11÷21.

Common pitfalls

• Using permutations instead of combinations; order does not matter for selection probabilities.
• Treating draws as independent with replacement; here they are dependent without replacement.
• Forgetting that the event is both draws non-blue (not just the first).

Final Answer10/21