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*}