Error in solution for Exercise 24.6.10

[aglebov]

The probabilities in the suggested $\pi$ do not add up to 1:

$$\begin{split} \frac{1-p}{p^4(1-p^5)} + \frac{1-p}{p^3(1-p^5)} + \frac{1-p}{p^2(1-p^5)} + \frac{1-p}{p(1-p^5)} + \frac{1-p}{1-p^5} &= \frac{1-p}{1-p^5} \left(\frac{1}{p^4} + \frac{1}{p^3} + \frac{1}{p^2} + \frac{1}{p} + 1\right) \\\ &= \frac{1-p}{1-p^5} \left(\frac{1 + p + p^2 + p^3 + p^4}{p^4}\right) \\\ &= \frac{1}{1 + p + p^2 + p^3 + p^4} \left(\frac{1 + p + p^2 + p^3 + p^4}{p^4}\right) \\\ &= \frac{1}{p^4} \\\ &\neq 1. \end{split}$$

Requiring $\pi_1 + \pi_2 + \pi_3 + \pi_4 + \pi_5 = 1$ yields

$$\pi = \begin{pmatrix} \frac{1-p}{1-p^5} & p \frac{1-p}{1-p^5} & p^2 \frac{1-p}{1-p^5} & p^3 \frac{1-p}{1-p^5} & p^4 \frac{1-p}{1-p^5} \end{pmatrix}$$