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 E₇ = event of getting an even number on one dice and a multiple of 3 on the other
Then E₇ = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6) , (3, 2), (6, 2), (3, 4), (6, 4), (3, 6)}
        i.e. n (E₇) = 11

\[\therefore P\left( E_7 \right) = \frac{n\left( E_7 \right)}{n\left( S \right)} = \frac{11}{36}\]