Two dice,one blue and one grey,are thrown at the same time.
$(i)$ Complete the following table:
$\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Sum} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \text{Prob.} & \frac{1}{36} & & & & & & \frac{5}{36} & & & & \frac{1}{36} \\ \hline \end{array}$
$(ii)$ $A$ student argues that 'there are $11$ possible outcomes $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ and $12$. Therefore,each of them has a probability $\frac{1}{11}$.' Do you agree with this argument? Justify your answer.

(N/A) $(i)$ When two dice are thrown,the total number of possible outcomes is $6 \times 6 = 36$. The sums and their corresponding probabilities are calculated as follows:
- Sum $= 2$: $(1,1) \rightarrow \frac{1}{36}$
- Sum $= 3$: $(1,2), (2,1) \rightarrow \frac{2}{36}$
- Sum $= 4$: $(1,3), (3,1), (2,2) \rightarrow \frac{3}{36}$
- Sum $= 5$: $(1,4), (4,1), (2,3), (3,2) \rightarrow \frac{4}{36}$
- Sum $= 6$: $(1,5), (5,1), (2,4), (4,2), (3,3) \rightarrow \frac{5}{36}$
- Sum $= 7$: $(1,6), (6,1), (2,5), (5,2), (3,4), (4,3) \rightarrow \frac{6}{36}$
- Sum $= 8$: $(2,6), (6,2), (3,5), (5,3), (4,4) \rightarrow \frac{5}{36}$
- Sum $= 9$: $(3,6), (6,3), (4,5), (5,4) \rightarrow \frac{4}{36}$
- Sum $= 10$: $(4,6), (6,4), (5,5) \rightarrow \frac{3}{36}$
- Sum $= 11$: $(5,6), (6,5) \rightarrow \frac{2}{36}$
- Sum $= 12$: $(6,6) \rightarrow \frac{1}{36}$
$(ii)$ No,$I$ do not agree with the student. The outcomes $2, 3, \dots, 12$ are not equally likely because the number of ways to obtain each sum varies. For example,there is only $1$ way to get a sum of $2$,but $6$ ways to get a sum of $7$.