Two fair six-faced dice are rolled simultaneously. What is the probability that the product of the numbers obtained is divisible by 4 but not divisible by 3?

ATQ, Total outcomes = 6 × 6 = 36. Count outcomes where product is divisible by 4. List by first die: 1: (1,4) 2: (2,2), (2,4), (2,6) 3: (3,4) 4: (4,1),(4,2),(4,3),(4,4),(4,5),(4,6) 5: (5,4) 6: (6,2),(6,4),(6,6) Total = 1 + 3 + 1 + 6 + 1 + 3 = 15. From these, remove outcomes where product is also divisible by 3 (i.e., divisible by 12). This happens if at least one die is 3 or 6. Mark multiples of 3 among the 15: (2,6), (3,4), (4,3), (4,6), (6,2), (6,4), (6,6) → 7 outcomes. Valid outcomes (divisible by 4 but not by 3) = 15 − 7 = 8. Probability = 8/36 = 2/9.