A bag contains (a + 2) red pens, (a – 3) green pens, and (a – 7) blue pens. Two pens are drawn in succession without replacement. The probability of drawing one red and one green pen is equal to the probability of rolling a total of at least 8 when two dice are thrown. Determine the probability of drawing a red pen from the bag.

Two dice are rolled: Total possible outcomes = 6 × 6 = 36 Favourable outcomes = (2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) → 15 outcomes. Therefore, the probability of getting a sum of 8 or more = 15/36 = 5/12. Number of pens in the bag = (a + 2) + (a – 3) + (a – 7) = 3a – 8. From the given condition: {^(a+2)C1 × ^(a-3) C1 } / ^(3a-8)C2 = 5/12

2(a + 2)(a – 3) × 12 = 5(3a – 8)(3a – 9) Or, 2(a + 2)(a – 3) × 12 = 15(3a – 8)(a – 3) Or, 2(a + 2) × 4 = 5(3a – 8) Or, 8a + 16 = 15a – 40 Or, 7a = 56 So, a = 8 Then: Total pens = 3 × 8 – 8 = 16 Red pens = 8 + 2 = 10 Probability (red) = 10/16 = 5/8