Question

Three six faced dice are thrown together. The probability that sum of the numbers appearing on the dice is equal to $k(3\le k\le 4)$ is

• A. $\frac{k^2}{108}$
• B. $\frac{k(k-1)}{216}$
• C. $\frac{(k-1)(k-2)}{2\times 6^3}$
• D. None of these

Answer 1

The correct option is C $\frac{(k-1)(k-2)}{2\times 6^3}$

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