Question

Three natural numbers are taken at random from the set $A=\{x|1\le x\le 100,xϵN\}$. The probability that the AM of the numbers is $25$, is

• A. $\frac{{}^{77}{C}_2}{{}^{100}{C}_3}$
• B. $\frac{{}^{25}{C}_2}{{}^{100}{C}_3}$
• C. $\frac{{}^{74}{C}_{72}}{{}^{100}{C}_{97}}$
• D. none of these

Answer 1

The correct option is C $\frac{{}^{74}{C}_{72}}{{}^{100}{C}_{97}}$

$n\left(S\right){=}^{100}{C}_3$
Since the $A.M.$ of three numbers is $25$.
Therefore their sum = $75$
$\therefore n\left(E\right)=$ the number of integral solution of $x_1+x_2+x_3=75$
Where $x_1\ge 1,x_2\ge 1,x_3\ge 1$
= coeff. of $x^{75}$ in ${(x+x^2+x^3+...)}^3$
=coeff. of $x^{72}$ in ${(1+x+x^2+...)}^3$
=coeff. of $x^{72}$ in ${\left(\frac{1}{1-x}\right)}^3$
=coeff. of $x^{72}$ in ${(1-x)}^{-3}$
${=}^{74}{C}_{72}{=}^{74}{C}_2$
$\therefore P\left(E\right)=\frac{{}^{74}{C}_2}{{}^{100}{C}_3}$