JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The probability that the first pair drawn consists of one blue and one green ball is: P_1 = ^ 6 C_ 1 ^ 6 C_ 1 ^ 12 C_ 2 = 36 66 = 6 11 Since the balls are drawn without replacement, 5 blue and 5 green balls remain. The probability that the second pair consists of one blue and one green ball is: P_2 = ^ 5 C_ 1 ^ 5 C_ 1 ^ 10 C_ 2 = 25 45 = 5 9 Continuing this process for all 6 pairs, the probabilities are: P_3 = ^ 4 C_ 1 ^ 4 C_ 1 ^ 8 C_ 2 = 16 28 = 4 7 P_4 = ^ 3 C_ 1 ^ 3 C_ 1 ^ 6 C_ 2 = 9 15 = 3 5 P_5 = ^ 2 C_ 1 ^ 2 C_ 1 ^ 4 C_ 2 = 4 6 = 2 3 P_6 = ^ 1 C_ 1 ^ 1 C_ 1 ^ 2 C_ 2 = 1 1 = 1 The required probability is the product of these individual probabilities: P = P_1 P_2 P_3 P_4 P_5 P_6 P = 6 11 5 9 4 7 3 5 2 3 1 Canceling common factors in the numerator and denominator: P = 6 4 2 11 9 7 = 48 693 = 16 231 Answer: 16 231