A fair dice is thrown twice. Find the probability that the sum obtained will be a factor of 12.
-
\(\frac{1}{3}\)
- \(\frac{5}{18}\)
- \(\frac{1}{9}\)
- \(\frac{7}{18}\)
The correct answer is: A
Explanation
When a die is thrown twice, the total number of possible outcomes is: 6 x 6 = 36
The factors of 12 are:1, 2, 3, 4, 6, 12
The possible sums from rolling two dice range from 2 (1 + 1) to 12 (6 + 6). We need to identify which sums are factors of 12:
Sum = 2: (1, 1)
Sum = 3: (1, 2), (2, 1)
Sum = 4: (1, 3), (2, 2), (3, 1)
Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
Sum = 12: (6, 6)
Total favorable outcome 1 + 2 + 3 + 5 + 1 = 12
The probability P that the sum is a factor of 12 is given by: \(\frac{\text{Favourable outcomes}}{\text{Total otcomes}}\) = \(\frac{12}{36}\) = \(\frac{1}{3}\)