Question

Three dice are thrown simultaneously. The probability of getting a sum of $15$ is

• A. $\frac{5}{108}$
• B. $\frac{1}{72}$
• C. $\frac{5}{72}$
• D. $\frac{5}{36}$

Answer 1

Answer: (A)

$n\left(S\right)=6^3=216$
$n\left(E\right)=$ sum of three numbers $x+y+z=15$
where$1\le x\le 6,1\le y\le 6,1\le z\le 6$
$n\left(E\right)=$ Coefficient of $x^{15}$ in ${\left(x+x^2+....+x^6\right)}^3$
= Coefficient of $x^{12}$ in ${\left(1+x+...+x^5\right)}^3$
= Coefficient of $x^{12}$ in ${\left(\frac{1-x^6}{1-x}\right)}^3$
= Coefficient of $x^{12}$ in ${\left(1-x^6\right)}^3{\left(1-x\right)}^{-3}$
= Coefficient of $x^{12}$ in $\left(1-3x^6+3x^{12}-x^{18}\right)\times$
$\left({}^2{C}_0{x}^0+{}^3{C}_1{x}^1+{}^4{C}_2{x}^2+...+{}^8{C}_6{x}^6+...+{}^{14}{C}_{12}{x}^{12}\right)$
$=3+\frac{8}{2}\times \frac{7}{1}\times \left(-3\right)+\frac{14}{2}\times \frac{13}{1}$
$=3-84+91=10$
$\therefore$ Required probability $=\frac{10}{216}=\frac{5}{108}$
Short Cut Method: Number of cases in which sum of three numbers $=15$ are
$\begin{array}{ccc}\left(5,5,5\right),& \left(6,6,3\right),& \left(6,4,5\right)\\ 1case& 3case& 6cases\end{array}$
$\therefore$ Required number of favourable cases are $1+3+6=10$
$\therefore$ Required probability$=\frac{10}{216}=\frac{5}{108}$