Question

Three six-faced fair dice are thrown together let P(k) denote the probability that the sum of the numbers appearing on the dice is $k(9\le k\le 4)$. Find 54S where $S=\sum _{k=9}^{14}P\left(k\right)$

Answer 1

The total number of cases is $6\times 6\times 6\times =6^3=216$. 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 $(1-3x^6)(1{+}^3{C}_{1}x{+}^4{C}_2{x}^2{+}^5{C}_3{x}^3+...)$
[note that $6\le k-3\le 11]$
${=}^{k-3+2}{C}_{k-3}-\left(3\right){}^{k-3+2-6}{C}_{k-3-6}$
${=}^{k-1}{C}_2-\left(3\right){}^{k-7}{C}_2$
Now $P\left(k\right)=\frac{21k-k^2-83}{216}$
$\therefore S=\frac{35}{54}$.