GIVEN : Two dices are taken. On faces of first dice, numbers from 1 to 6 are written. On second dice, six consecutive natural numbers are written. CALCULATION : If there are six consecutive natural numbers on second dice, then exactly two of them will be multiple of 3, exactly two will leave a remainder of 1 when divided by 3, and exactly two will leave a remainder of 2 when divided by 3. The sum of outcomes when divided by 3 will leave a remainder of 1 when the sum of remainders of individual outcomes is either 1 or 4. This is possible when remainders of outcomes when divided by 3 are (0, 1), (1, 0), (2, 2). For (0,1), outcome of first dice can be 3 or 6, and outcome of second dice can be a number that will leave a remainder of 1 when divided by 3. This gives a total of 4 cases. For (1,0), outcome of first dice can be 1 or 4, and outcome of second dice can be a number that will leave a remainder of 0 when divided by 3. This gives a total of 4 cases. For (2,2), outcome of first dice can be 2 or 5, and outcome of second dice can be a number that will leave a remainder of 2 when divided by 3. This gives a total of 4 cases. So, there are 12 favourable cases. ∴ Required probability = 12/(6 × 6) = 1/3