Mathematics - Probability Question with Solution | TestHub

$\frac{{\sum }_{i=1}^n{i}^2}{{\sum }_{i=1}^{n}i}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}$ For $n=1,2,3,\dots ..,1000$ Value of $\frac{2n+1}{3}=\frac{3}{3},\frac{5}{3},\frac{7}{3},\dots \dots \dots ,\frac{2001}{3}$ respectively. Out of $\frac{3}{3},\frac{5}{3},\frac{7}{3},\dots \dots \dots ,\frac{2001}{3}$ only first term $\left(\frac{3}{3}=1\right)$, fourth term $\left(\frac{9}{3}=3\right),667^{\text{th }}$ term $\left(\frac{2001}{3}=667\right)$ are integers. Hence, out of 1000 values of $\frac{2n+1}{3}$, total number of integral values of $\frac{2n+1}{3}$ $=333+1=334$ $\therefore$ Required probability $=\frac{334}{1000}=0.334$