Question

Three six-faced fair dice are thrown together. The probability that the sum of the numbers appearing on the dice is k $(3\le k\le 8)$ is

• A. $\frac{(k-1)(k-2)}{132}$
• B. $\frac{k(k-1)}{432}$
• C. $\frac{k^2}{132}$
• D. $\frac{(k+1)(k+2)}{432}$

Answer 1

The correct option is A $\frac{(k-1)(k-2)}{132}$

Total number of ways = $6^3$ =216
Number of favourable cases =
number of positive integral solutions of $x_1+x_2+x_3=k,3\le k\le 8$
= coefficient of $x^{k}in(x+x^2+...+x^6{)}^3=coefficientof{x}^{k-3}in(1+x+.....+x^5{)}^3$
= coefficient of $x^{k-3}in(1-x^6{)}^3(1-x{)}^{-3}since3\le k\le 8\Rightarrow \le k-3\le 5$
$=(k-3+3-1)C_2=(k-1)C_2=$
Hence required probability = $\frac{(k-1)(k-2)}{2\times 216}$