Question

Three six faced 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. $\frac{k(k-1)}{432}$
• B. $\frac{k^2}{432}$
• C. $\frac{(k-1)(k-2)}{432}$
• D. None of these

Answer 1

The correct option is C $\frac{(k-1)(k-2)}{432}$

Total number of ways $=6^3=216$
Number of favorable ways $=$ coefficients of $x^k$ in ${(x+x^2+...+x^6)}^3$
$=$ coefficients of $x^{k-3}$ in ${(1+x+x^2+...+x^6)}^3$
$=$ coefficients of $x^{k-3}$ in ${(1-x^6)}^3{(1-x)}^{-3}$
Since $3\le k\le 8\Rightarrow 0\le k-3\le 5$, we have numner of favorable ways
$=$ coefficients of $x^{k-3}$ in ${(1-x)}^{-3}$
${=}^{k-3+3-1}{C}_{k-3}{=}^{k-1}{C}_2$
$\therefore$ Required probability $=\frac{{}^{k-1}{C}_2}{216}=\frac{(k-1)(k-2)}{432}$