Two dice are thrown simultaneously.

Two dice are thrown simultaneously.                                                                                      [given]

So, total number of possible outcomes = 36

(i) Sum of the numbers appearing on the dice is $7 .$

So, the possible ways are $(1,6),(2,5),(3,4),(4,3),(5,2)$ and $(6,1)$.

Number of possible ways $=6$

$\therefore \quad$ Required probability $=\frac{6}{36}=\frac{1}{6}$

(ii) Sum of the numbers appearing on the dice is a prime number $i, e, 2,3,5,7$ and 11 .

So, the possible ways are $(1,1),(1,2),(2,1),(1,4),(2,3),(3,2),(4,1),(1,6),(2,5)$, $(3,4),(4,3),(5,2),(6,1),(5,6)$ and $(6,5)$.

Number of possible ways $=15$

$\therefore$ Required probability $=\frac{15}{36}=\frac{5}{12}$

(iii) Sum of the numbers appearing on the dice is $1 .$ It is not possible, so its probability is zero.