A box contains ‘x + 2’ red, ‘x – 1’ green, and ‘x – 3’ blue balls. If two balls are drawn from the box one after another without replacement, the probability of getting a red ball and a green ball is the same as the probability of getting a sum divisible by 3 when two dice are rolled. Find the probability of getting a red ball from the box. 

When two dice are rolled: Total outcomes = 6 × 6 = 36 Favourable outcomes for a sum divisible by 3 (sum = 3, 6, 9, or 12): • Sum = 3: (1, 2), (2, 1) = 2 outcomes • Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) = 5 outcomes • Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) = 4 outcomes • Sum = 12: (6, 6) = 1 outcome Total favourable outcomes = 2 + 5 + 4 + 1 = 12 So, probability of getting a sum divisible by 3 = 12/36 = 1/3 Total number of balls in the box = (x + 2) + (x – 1) + (x – 3) = 3x – 2 According to question: {(x + 2)C1 × (x – 1)C1} / (3x – 2)C2 = 1/3 Or, {2(x + 2)(x – 1)} / {(3x – 2)(3x – 3)} = 1/3 Or, {2(x + 2)(x – 1)} / {3(3x – 2)(x – 1)} = 1/3 Or, {2(x + 2)} / {3(3x – 2)} = 1/3 Or, 2(x + 2) = 3x – 2 Or, 2x + 4 = 3x – 2 Or, x = 6 Total number of balls = 3(6) – 2 = 16 Number of red balls = 6 + 2 = 8 Desired probability = 8/16 = 1/2