Question

Three dices 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

The correct option is A $\frac{5}{108}$

$n\left(S\right)=6^3=216$
$n\left(E\right)=$ sum of three number $x+y+z=15$
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 $(x+x^2+....+x^6{)}^3$
$=$ Coefficient of $x^{12}$ in $(1+x+...+x^5{)}^3$
$=$ Coefficient of $x^{12}$ in ${\left(\frac{1-x^6}{1-x}\right)}^3$
$=$ Coefficient of $x^{12}$ in ${(1-x^6)}^3{(1-x)}^{-3}$
$=$ Coefficient of $x^{12}(1-3x^6+3x^{12}-x^{18})$
$({}^2{\text{C}}_0{x}^0{+}^3{\text{C}}_1{x}^1{+}^4{\text{C}}_2{x}^2+\cdots {+}^8{\text{C}}_6{x}^6+\cdots {+}^{14}{\text{C}}_{12}{x}^{12})$
${=}^2{\text{C}}_0\times 3{+}^8{\text{C}}_6\times (-3){+}^{14}{\text{C}}_{12}$
$=3+\frac{8}{2}\times \frac{7}{1}\times (-3)+\frac{14}{2}\times \frac{13}{1}$
$=3-84+91$
$=10$
$\therefore$ Required probability $=\frac{10}{216}=\frac{5}{108}$