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

Answer 1

The correct option is A

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

Explanation for the correct answer:

Since three die are thrown, the total number of outcomes are $6^3=216$

Now, for the favorable number of outcomes we consider the coefficient of $x^k$ in ${\left(x+x^2+x^3+....+x^6\right)}^3$

Multiplying and dividing by ${\left(1-x\right)}^3$ we get

$\Rightarrow$ coefficient of $x^k$ in ${\left(\frac{\left(1-x\right)\left(x+x^2+x^3+...+x^6\right)}{1-x}\right)}^3$

$\Rightarrow$ coefficient of $x^k$ in ${\left(\frac{x-x^7}{1-x}\right)}^3$

$\Rightarrow$ coefficient of $x^k$ in $x^3{\left(\frac{1-x^6}{1-x}\right)}^3$

$\Rightarrow$ coefficient of $x^{k-3}$ in ${\left(\frac{1-x^6}{1-x}\right)}^3$ …[Dividing by $x^3$ throughout]

$\Rightarrow$ coefficient of $x^{k-3}$ in $\left(1-x^6\right){\left(1-x\right)}^{-3}$

$\Rightarrow$ coefficient of $x^{k-3}$ in ${\left(1-x\right)}^{-3}$ $...\left[\because 0\le k-3\le 5\right]$

$\Rightarrow$ 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}_{3-1}$

$=\frac{\left(k-1\right)!}{2!\left(k-3\right)!}$

$=\frac{\left(k-1\right)\left(k-2\right)}{2}$

Hence, the $\mathbf{Required}\mathbf{}\mathbf{probability}\mathbf{=}\mathbf{}\frac{\mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{favorable}\mathbf{}\mathbf{outcomes}}{\mathbf{Total}\mathbf{}\mathbf{number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{outcomes}}$

$\Rightarrow$ $\mathrm{Required}\mathrm{probability}=\frac{\left(k-1\right)\left(k-2\right)}{2\times 216}$

$\Rightarrow$ $\mathrm{Required}\mathrm{probability}=\frac{\left(k-1\right)\left(k-2\right)}{432}$

Hence, the required probability is $\frac{\left(k-1\right)\left(k-2\right)}{432}$.

Hence, option (A) is the correct answer.