Answer to Question #323769 in Statistics and Probability for ash

Question

A random variable has PDF p_i = 2(1/3)^i and takes on values in the positive integers. What is the probability that it takes on a value between 10 and 20, inclusive? Show your work step by step.

Answer 1

$P\left( 10\leqslant X\leqslant 20 \right) =\sum_{n=10}^{20}{P\left( X=n \right)}=\\=\sum_{n=10}^{20}{2\cdot \left( \frac{1}{3} \right) ^n}=\left[ \begin{array}{c} geometric\,\,progression\\ with\,\,11 terms, b_0=\left( \frac{1}{3} \right) ^{10},q=\frac{1}{3}\\\end{array} \right] =\\=2\frac{\left( \frac{1}{3} \right) ^{10}\left( 1-\left( \frac{1}{3} \right) ^{12} \right)}{1-\frac{1}{3}}=\frac{1}{3^9}\left( 1-\frac{1}{3^{12}} \right) =5.08052\times 10^{-5}$