Two dice,one blue and one grey,are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
$(i)$ $8 ?$
$(ii)$ $13 ?$
$(iii)$ less than or equal to $12 ?$

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(N/A) When the blue die shows $1$,the grey die could show any one of the numbers $1, 2, 3, 4, 5, 6$. The same is true when the blue die shows $2, 3, 4, 5,$ or $6$. The possible outcomes of the experiment are listed in the table below; the first number in each ordered pair is the number appearing on the blue die and the second number is that on the grey die.
Note that the pair $(1, 4)$ is different from $(4, 1)$.
So,the total number of possible outcomes $= 6 \times 6 = 36$.
$(i)$ The outcomes favourable to the event "the sum of the two numbers is $8$" denoted by $E$,are: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$.
i.e.,the number of outcomes favourable to $E = 5$.
Hence,$P(E) = \frac{5}{36}$.
$(ii)$ As you can see,there is no outcome favourable to the event $F$,"the sum of two numbers is $13$".
So,$P(F) = \frac{0}{36} = 0$.
$(iii)$ As you can see,all the outcomes are favourable to the event $G$,"sum of two numbers $\leq 12$".
So,$P(G) = \frac{36}{36} = 1$.

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