Question

The probability of getting a sum of $12$ in four throws of an ordinary dice is

• A. $\frac{1}{6}{\left(\frac{5}{6}\right)}^3$
• B. ${\left(\frac{5}{6}\right)}^4$
• C. $\frac{1}{36}{\left(\frac{5}{6}\right)}^2$
• D. None of these

Answer 1

The correct option is A $\frac{1}{6}{\left(\frac{5}{6}\right)}^3$

$n\left(S\right)=6\times 6\times 6\times 6$.

$n\left(E\right)=$ the number of integral solutions of $x_1+x_2+x_3+x_4=12$,

where $1\le x_1\le 6,...1\le x_4\le 6$

$=$ coefficient of $x^{12}$ in ${(x+x^2+...+x^6)}^4$

$=$ coefficient of $x^8$ in ${\left(\frac{1-x^6}{1-x}\right)}^4$

$=$ coefficient of $x^8$ in ${(1-x^6)}^4\times \left({}^3{C}_0{+}^4{C}_{1}x{+}^5{C}_2{x}^2+...\right)$

${=}^{11}{C}_8-4^5{C}_2=125$

$\therefore P\left(E\right)=\frac{125}{6\times 6\times 6\times 6}$