Question

Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3

Answer 1

We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36

Let E₂ be the event of getting the sum of the numbers obtained on the two dice is neither a multiple of 2 nor a multiple of 3. Then,
E₂' = event of getting the sum of the numbers obtained on the two dice is either a multiple of 2 or a multiple of 3.
∴ E₂' = {(1,1), (1,2), (2, 1), (1, 3), (2, 2), (3, 1), (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6),
            (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (6, 6)}
i.e. n(E₂') = 24
Thus, P(E₂') = \[\frac{24}{36} = \frac{2}{3}\]

Hence, required probability P(E₂) = 1 - P(E₂')

\[= 1 - \frac{2}{3} = \frac{3 - 2}{3} = \frac{1}{3}\]