Chapter 2 Section 3 Question 9
Question
Let $Z$ ~ $Negative-Binomial(3, 1/4)$. Compute $P(Z \leq 2)$.
From Example 2.3.5 in the book we know: $$p\scriptsize{Z}\normalsize{(z)=P(Z=z)={{r-1+z}\choose z}\theta^r(1-\theta )^z}$$ In this equation r represents the desired number of successes, z represents the number of failures, $\theta$ represents the probability of a success.
we shall now plug in $\theta=\frac{1}{4}$, $r=3$ $$P(Z=z)={{3-1+z}\choose z}(\frac{1}{4})^3(1-\frac{1}{4} )^z$$ $$P(Z=z)={{2+z}\choose z}(\frac{1}{4})^3(\frac{3}{4} )^z$$ To solve for $Z\leq 2$ we will add up z=0,1,2 together $$P(Z\leq 2)={{2+0}\choose 0}(\frac{1}{4})^3(\frac{3}{4} )^0+{{2+1}\choose 1}(\frac{1}{4)}^3(\frac{3}{4} )^1+{{2+2}\choose 2}\frac{1}{4}^3(\frac{3}{4} )^2$$ $$P(Z\leq 2)={{2}\choose 0}(\frac{1}{4})^3(\frac{3}{4} )^0+{{3}\choose 1}(\frac{1}{4})^3(\frac{3}{4} )^1+{{4}\choose 2}(\frac{1}{4})^3(\frac{3}{4} )^2$$ $$P(Z\leq 2)=(\frac{1}{4})^3(\frac{3}{4} )^0+3*(\frac{1}{4})^3(\frac{3}{4} )^1+6*(\frac{1}{4})^3(\frac{3}{4} )^2$$ $$P(Z\leq 2)=(\frac{1}{4})^3+3*\frac{1}{64}(\frac{3}{4} )+6*\frac{1}{64}(\frac{9}{16} )$$ $$P(Z\leq 2)=\frac{1}{64}+3*\frac{3}{256}+6*\frac{9}{1024}$$ $$P(Z\leq 2)=\frac{53}{512}$$