Solution 8.3.9.1.

Using the exponential distribution,
\begin{equation*} p = P(X \gt 1) = 1 - F(1) = 1 - (1-e^{-\frac{1}{0.7}}) \approx 0.2397. \end{equation*}
Now, use the geometric distribution with \(p = 0.2397\text{...}\)
\begin{equation*} P(X = 5) = f(5) = (1-0.2397)^4 \cdot 0.2397 \approx 0.0801. \end{equation*}
\begin{equation*} P(\text{2nd success on 8} | \text{1st success on 3}) = P(X = 5) \approx 0.0801 \end{equation*}
\begin{equation*} \begin{aligned} P(\text{1st success on an odd trial}) \amp = f(1) + f(3) + f(5) + ...\\ \amp = p + (1-p)^2 \cdot p + (1-p)^4 \cdot p + (1-p)^6 \cdot p + ... \\ \amp = p \sum_{x=0}^{\infty} \left ( (1-p)^2 \right)^k \\ \amp = p \frac{1}{1-(1-p)^2} \approx 0.5681 \end{aligned} \end{equation*}
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