Question

Two dice are thrown find the probability of getting
(i) The sum of the numbers on their upper faces is divisible by 9.
(ii) The sum of the numbers on their upper faces is at the most 3.
(iii) The number on the upper face of the first die is less than the number on the upper face of the second die.

Answer 1

Given:
Two dice are thrown.
Sample space = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Or,
n(S) = 36

(i) Let A be the event that the sum of the numbers on the upper faces of the dice is divisible by 9.
Here,
A = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(A) = 4
Thus, we have:
P(A) =$\frac{n\left(\mathrm{A}\right)}{n\left(\mathrm{S}\right)}$
$\Rightarrow$P(A) =$\frac{4}{36}=\frac{1}{9}$

(ii) Let B be the event that the sum of the numbers on the upper faces of the dice is at most 3.
Here,
B = {(1, 1), (1, 2), (2, 1)}
∴ n(B) = 3
Thus, we have:
$\mathrm{P}\left(\mathrm{B}\right)=\frac{n\left(\mathrm{B}\right)}{n\left(\mathrm{S}\right)}$
$\Rightarrow \mathrm{P}(\mathrm{B})=\frac{3}{36}=\frac{1}{12}$

(iii) Let C be the event that the number on the upper face of the first die is less than the number on the upper face of the second die.
Here,
C = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6)}

∴ n(C) = 15
Thus, we have:
$$\begin{gathered} \mathrm{P}\left(\mathrm{C}\right)=\frac{n\left(\mathrm{C}\right)}{n\left(\mathrm{S}\right)} \\ \Rightarrow \mathrm{P}\left(\mathrm{C}\right)=\frac{15}{36}=\frac{5}{12} \end{gathered}$$