Question

Three six-faced dice are thrown together. The probability that the sum of the numbers appearing on the dice is $k(9\le k\le 14)$ is

• A. $\frac{21k-k^2-83}{216}$
• B. $\frac{k^2-3k+2}{432}$
• C. $\frac{21k-k^2-83}{432}$
• D. None of these

Answer 1

The correct option is A $\frac{21k-k^2-83}{216}$

We have, favorable number of elementary events
$=$ coefficients of $x^{k-3}$ in ${(1-x^6)}^3{(1-x)}^{-3}$
$=$ coefficients of $x^{k-3}$ in $(1{-}^3{C}_1{x}^6+...){(1-x)}^{-3}$ $[\because 9\le k\le 14,\therefore 6\le k-3\le 11]$
$=$ coefficients of $x^{k-3}$ in ${(1-x)}^{-3}{-}^3{C}_1($ coefficients of $x^{k-9}$ in ${(1-x)}^{-3})$
${=}^{k-3+3-1}{C}_{3-1}{-}^3{C}_1{\times }^{k-9+3-1}{C}_{3-1}{=}^{k-1}{C}_2-{3.}^{k-7}{C}_2$
$=\frac{(k-1)(k-2)}{2}-3.\frac{(k-7)(k-8)}{2}=21k-k^2-83$
Hence, the probability of the required event
$=\frac{21k-k^2-83}{6^3}$