Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
$(i)$ $7 ?$
$(ii)$ a prime number?
$(iii)$ $1 ?$

(A-D) When two dice are thrown simultaneously,the total number of possible outcomes is $6 \times 6 = 36$.
$(i)$ Sum of the numbers appearing on the dice is $7$. The favorable outcomes are $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$.
Number of favorable outcomes $= 6$.
$\text{Probability} = \frac{6}{36} = \frac{1}{6}$.
$(ii)$ Sum of the numbers appearing on the dice is a prime number. The possible prime sums are $2, 3, 5, 7, 11$.
- Sum $= 2: (1, 1) \rightarrow 1$ outcome
- Sum $= 3: (1, 2), (2, 1) \rightarrow 2$ outcomes
- Sum $= 5: (1, 4), (2, 3), (3, 2), (4, 1) \rightarrow 4$ outcomes
- Sum $= 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) \rightarrow 6$ outcomes
- Sum $= 11: (5, 6), (6, 5) \rightarrow 2$ outcomes
Total favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.
$\text{Probability} = \frac{15}{36} = \frac{5}{12}$.
$(iii)$ Sum of the numbers appearing on the dice is $1$. Since the minimum sum of two dice is $1 + 1 = 2$,it is impossible to get a sum of $1$.
Number of favorable outcomes $= 0$.
$\text{Probability} = \frac{0}{36} = 0$.