Question

A pair of $12$-sided fair dice with faces numbered $1,2,3,...,12$ is rolled. The probability that the sum of the numbers appearing has remainder $2$ when divided by $9$ is?

• A. $\frac{7}{72}$
• B. $\frac{5}{48}$
• C. $\frac{11}{144}$
• D. $\frac{1}{9}$

Answer 1

Answer: (D)

$\frac{1}{9}$

Let $a_1$ be the number appearing on dice 1.

Let $a_2$ be the number appearing on dice 2.

$1\le a_1\le 12$

$1\le a_2\le 12$

Let $S$ be the sum of number on both dice.

$S=a_1+a_2$ ...(1)

Given that $s$ should be of the form $9r+2$

From (1), $9r+2=a_1+a_2$

For $r=0$ $\Rightarrow$ no of case $=1$

Sr. No.	$S$	$a_1$	$a_2$
1	2	1	1

For $r=1$ $\Rightarrow$ no of case $=10$

Sr. No.	$S$	$a_1$	$a_2$
1	11	1	10
2	11	10	1
3	11	2	9
4	11	9	2
5	11	3	8
6	11	8	3
7	11	4	7
8	11	7	4
9	11	5	6
10	11	6	5

For $r=2$ $\Rightarrow$ no of case $=5$

Sr. No.	$S$	$a_1$	$a_2$
1	20	8	12
2	20	12	8
3	20	9	11
4	20	11	9
5	20	10	10

Total no of favourable case $=1+5+10=16$

Total no of case $=12\times 12=144$

$\therefore$ Probability of required case $=\frac{16}{144}=\frac{1}{9}$