Question:
Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i) 6
(ii) 12
(iii) 7
Solution:
Number of total outcomes = 36
(i) When product of the numbers on the top of the dice is $6 .$
So, the possible ways are $(1,6),(2,3),(3,2)$ and $(6,1)$.
Number of possible ways $=4$
$\therefore \quad$ Required probability $=\frac{4}{36}=\frac{1}{9}$
(ii) When product of the numbers on the top of the dice is $12 .$ So, the possible ways are $(2,6),(3,4),(4,3)$ and $(6,2)$. Number of possible ways $=4$
$\therefore \quad$ Required probability $=\frac{4}{36}=\frac{1}{9}$
(iii) Product of the numbers on the top of the dice cannot be $7 .$ So, its probability is zero.