Chapter 2 Section 3 Question 10
Let $X$ ~ $Geometric(1/5)$. Compute $P(X^2 \leq 15)$.
Author: Mohammad-Ali Bandzar| Date:Oct 16 2020
From Example 2.3.4 in the book we know: $$p\scriptsize{X}\normalsize{(x)=P(X=x)=(1-\theta)^x\theta$$ In this equation x represents the number of failures before a successful trial, $\theta$ represents the probability of a success. (x+1) would represent the total number of trials

we shall now plug in $\theta=\frac{1}{5}$ $$P(X=x)=(1-\frac{1}{5})^x\frac{1}{5}$$ $$P(X=x)=(\frac{4}{5})^x\frac{1}{5}$$ We know that we want $P(X^2 \leq 15)$ $$X^2 \leq 15$$ $$X \leq\pm\sqrt{15}$$ since X must be greater than zero we can ignore the negative root $$X \leq\sqrt{15}$$ $$X \leq\approx 3.8729$$ we now want to round down to the nearest whole number as we must have an integer number of failures.(you cant fail .87 trials haha)
$$X \leq 3$$ We will now add up the probability for k=0,1,2,3 failures together $$P(X\leq 3)=(\frac{4}{5})^0\frac{1}{5}+(\frac{4}{5})^1\frac{1}{5}+(\frac{4}{5})^2\frac{1}{5}$$ $$P(X\leq 3)=\frac{1}{5}+(\frac{4}{5})\frac{1}{5}+(\frac{16}{25})\frac{1}{5}$$ $$P(X\leq 3)=\frac{1}{5}+(\frac{4}{25})+(\frac{16}{125})$$ $$P(X\leq 3)=\frac{1}{5}+(\frac{4}{25})+(\frac{16}{125})$$ $$P(X\leq 3)=\frac{61}{125}$$