Question

Five ordinary dies are rolled at random and the sum of numbers shown is $16$. What is the probability that the number shown on each is any one from $2,3,4$ or $5?$

• A. $\frac{9}{49}$
• B. $\frac{3}{49}$
• C. $\frac{2}{49}$
• D. None of these

Answer 1

The correct option is A $\frac{9}{49}$

If the number shown be integers $x_1,x_2,x_3,x_4$ and $x_5,$

then $x_1+x_2+x_3+x_4+x_5=16,$

where $1\le x_1\le 6;1\le x_2,\le 6;...;1\le x_5\le 6.$

The number of solutions of this equation

$=$coefficient of $x^{16}$ in ${(x+x^2+x^3+x^4+x^5+x^6)}^5$

$=$coefficient of $x^{11}$ in ${(1+x+x^2+x^3+x^4+x^5)}^5$

$=$coefficient of $x^{11}$ in ${\left(\frac{1-x^6}{1-x}\right)}^5=$coefficient of $x^{11}$ in ${(1-x^6)}^5.{(1-x)}^{-5}$

$=$coefficient of $x^{11}$ in $\{{}^5{C_0-}^5{C}_1.x^6{+}^5{C}_2{x}^{12}-...\}\times \{{}^4{C_0+}^5{C}_{1}x+...{+}^{15}{C}_{11}{x}^{11}+...+\infty \}$

$\{{}^5{C_0-}^5{C}_1.x^6{+}^5{C}_2{x}^{12}-...\}\times \{{}^4{C_0+}^5{C}_{1}x+...{+}^{15}{C}_{11}{x}^{11}+...+\infty \}$

${=}^5{C}_0{.}^{15}{C_{11}-}^5{C_1.}^9{C}_5=\frac{\mathrm{15.14.13.12.}}{24}-5\frac{\mathrm{9.8.7.6}}{24}$

$=\mathrm{15.7.13}-\mathrm{10.9.7}=1365-630=735$

$\therefore n\left(S\right)=735$

Now, $n\left(E\right)=$ the number of integral solutions of $x_1+x_2+...+x_5=16$

where $2\le x_1\le 5,2\le x_2\le 5,...,2\le x_5\le 5$

$=$coefficient of $x^{16}$ in ${(x^2+x^3+x^4+x^5)}^5$

$=$coefficient of $x^6$ in ${(1+x+x^2+x^3)}^3=$coefficient of $x^6$ in ${(1+x)}^5.{(1+x^2)}^5=$coefficient of $x^6$ in $\left\{{}^5{C_0{+}^5{C}_{1}x+}^5{C}_2.x^2+...{+}^5{C}_5.x^5\right\}\times \left\{{}^5{C}_0{+}^5{C}_1{x}^2{+}^5{C}_2{x}^4+...{+}^5{C}_5{x}^{10}\right\}$

${=}^5{C_0.}^5{C}_3{+}^5{C}_2{.}^5{C}_2{+}^5{C}_4{.}^5{C}_1$

$=10+10\times 10+5\times 5=135$

$\therefore$ the required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{135}{735}=\frac{9}{49}$