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

Answer 1

Answer: (A)

$\frac{(k-1)(k-2)}{432}$

The total number of cases is $6\times 6\times 6=6^3=216$. The number of favourable ways
$=$ coefficient of $x^k$ in $(x+x^2+...+x^6{)}^3$
$=$ coefficient of $x^{k-3}$ in $(1-x^6{)}^3(1-x{)}^{-3}$
$=$ coefficient of $x^{k-3}$ in $(1-x{)}^3[\because 0\le k-3\le 5]$
$=$ coefficient of $x^{k-3}$ in $(1{+}^3{C}_{1}x{+}^4{C}_2{x}^2{+}^5{C}_3{x}^3+.....)$
${=}^{k-3+2}{C}_{k-3}{=}^{k-1}{C}_2=\frac{(k-1)(k-2)}{2}$
Thus, the probability of the required event is $\frac{(k-1)(k-2)}{432}$.