$A$ die is numbered in such a way that its faces show the numbers $1, 2, 2, 3, 3, 6$. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:

+	$1$	$2$	$2$	$3$	$3$	$6$
$1$	$2$	$3$	$3$	$4$	$4$	$7$
$2$	$3$	$4$	$4$	$5$	$5$	$8$
$2$	$3$	$4$	$4$	$5$	$5$	$8$
$3$	$4$	$5$	$5$	$6$	$6$	$9$
$3$	$4$	$5$	$5$	$6$	$6$	$9$
$6$	$7$	$8$	$8$	$9$	$9$	$12$

What is the probability that the total score is
$(i)$ even?
$(ii)$ $6?$
$(iii)$ at least $6?$

(N/A) The total number of possible outcomes when two dice are thrown is $6 \times 6 = 36$.
$(i)$ The outcomes where the sum is even are:
$2, 2, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 8, 8, 8, 8, 12$ (counting all occurrences).
Total number of outcomes with an even sum $= 18$.
$P(\text{even sum}) = \frac{18}{36} = \frac{1}{2}$.
$(ii)$ The outcomes where the sum is $6$ are:
$(3, 3), (3, 3), (3, 3), (3, 3)$ (from the table).
Total number of outcomes with sum $6 = 4$.
$P(\text{sum is } 6) = \frac{4}{36} = \frac{1}{9}$.
$(iii)$ The outcomes where the sum is at least $6$ (i.e., $\ge 6$) are:
$6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 12$.
Total number of outcomes with sum at least $6 = 15$.
$P(\text{sum at least } 6) = \frac{15}{36} = \frac{5}{12}$.