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$1 \le x \le6, 1 \le y \le6, 1\le z \le 6$
$n\left(E\right) =$ 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
$\begin{matrix}\left(5, 5, 5\right),&\left(6, 6,3\right),&\left(6, 4, 5\right)\\ 1 case&3case&6 cases\end{matrix}$
$\therefore $ Required number of favourable cases are $1 + 3 + 6 = 10$
$\therefore $ Required probability$= \frac{10}{216} = \frac{5}{108}$