Probability & Statistics Ch.2 Section 3 Question 9

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Let $Z \sim \text{Negative-Binomial}(3, 1/4)$ . Compute $P(Z \leq 2)$ .

Solution

From Example 2.3.5 in the textbook, we know:

p_Z(z) = P(Z=z) = \binom{r-1+z}{z}\theta^r(1-\theta)^z

Where:

r = desired number of successes
z = number of failures
θ = probability of success

We plug in $\theta = \frac{1}{4}$ , $r = 3$ :

P(Z=z) = \binom{2+z}{z}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^z

To solve for $P(Z \leq 2)$ , we add up z = 0, 1, 2:

P(Z \leq 2) = \binom{2}{0}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^0 + \binom{3}{1}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^1 + \binom{4}{2}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2

= 1 \cdot \frac{1}{64} \cdot 1 + 3 \cdot \frac{1}{64} \cdot \frac{3}{4} + 6 \cdot \frac{1}{64} \cdot \frac{9}{16}

= \frac{1}{64} + \frac{9}{256} + \frac{54}{1024}

= \frac{16}{1024} + \frac{36}{1024} + \frac{54}{1024}

P(Z \leq 2) = \frac{106}{1024} = \frac{53}{512}