Two dice are numbered $1, 2, 3, 4, 5, 6$ and $1, 1, 2, 2, 3, 3$ respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from $2$ to $9$ separately.

(N/A) Total number of outcomes $= 6 \times 6 = 36$.
$(i)$ Event $E_1$ (sum $= 2$): Outcomes are $\{(1, 1), (1, 1)\}$. $n(E_1) = 2$. $P(E_1) = \frac{2}{36} = \frac{1}{18}$.
$(ii)$ Event $E_2$ (sum $= 3$): Outcomes are $\{(1, 2), (1, 2), (2, 1), (2, 1)\}$. $n(E_2) = 4$. $P(E_2) = \frac{4}{36} = \frac{1}{9}$.
$(iii)$ Event $E_3$ (sum $= 4$): Outcomes are $\{(1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)\}$. $n(E_3) = 6$. $P(E_3) = \frac{6}{36} = \frac{1}{6}$.
$(iv)$ Event $E_4$ (sum $= 5$): Outcomes are $\{(2, 3), (2, 3), (3, 2), (3, 2), (4, 1), (4, 1)\}$. $n(E_4) = 6$. $P(E_4) = \frac{6}{36} = \frac{1}{6}$.
$(v)$ Event $E_5$ (sum $= 6$): Outcomes are $\{(3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)\}$. $n(E_5) = 6$. $P(E_5) = \frac{6}{36} = \frac{1}{6}$.
$(vi)$ Event $E_6$ (sum $= 7$): Outcomes are $\{(4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)\}$. $n(E_6) = 6$. $P(E_6) = \frac{6}{36} = \frac{1}{6}$.
$(vii)$ Event $E_7$ (sum $= 8$): Outcomes are $\{(5, 3), (5, 3), (6, 2), (6, 2)\}$. $n(E_7) = 4$. $P(E_7) = \frac{4}{36} = \frac{1}{9}$.
$(viii)$ Event $E_8$ (sum $= 9$): Outcomes are $\{(6, 3), (6, 3)\}$. $n(E_8) = 2$. $P(E_8) = \frac{2}{36} = \frac{1}{18}$.