Probability of catching the first bus:
\begin{gather*}
\textbf{P(A) = 0.6}
\end{gather*}
Probability of missing the first bus
\begin{gather*}
\textbf{P}(\textbf{A}^{c}) \, = \, 1 - \textbf{P(A)} \, = \, 1 - 0.6 \, = \, 0.4
\end{gather*}
Probability of catching the second bus, given that the first bus was missed
\begin{gather*}
\textbf{P}(\textbf{B}|\textbf{A}^{c}) \, = \, 0.8
\end{gather*}
Probability of missing the second bus, given that the first bus was missed
\begin{gather*}
\textbf{P}(\textbf{B}^{c} | \textbf{A}^{c} ) \, = \, 1 - \textbf{P}(\textbf{B} | \textbf{A}^{c} ) \, = \, 1 - 0.8 \, = \, 0.2
\end{gather*}
-
The probability of catching the first bus is directly given as
\begin{gather*}
\textbf{P(A)} \, = \, 0.6
\end{gather*}
The probability of catching the first bus is
\(0.6\) or
\(60\%\text{.}\)
-
Missing the first bus
\(\textbf{A}^{c}\)
Catching the second bus
\(\text{B}\)
Since these events are dependent, we use the multiplication rule
\begin{gather*}
\textbf{P}(\textbf{A}^{c} \cap \textbf{B} ) \, = \, \textbf{P}(\textbf{A}^{c}) \times \textbf{P} (\textbf{B} | \textbf{A}^{c})
\end{gather*}
\begin{gather*}
\textbf{P}(\textbf{A}^{c} \cap \textbf{B} ) \, = \, 0.4 \times 0.8 \, = \, 0.32
\end{gather*}
The probability of missing the first bus but catching the second is
\(0.32\) or
\(32\%\text{.}\)
-
Missing the first bus
\(\textbf{A}^{c}\)
Missing the second bus
\(\textbf{B}^{c}\)
Again, using the multiplication rule
\begin{gather*}
\textbf{P}(\textbf{A}^{c} \cap \textbf{B}^{c} ) \, = \, \textbf{P(A)}^{c} \times \textbf{P}(\textbf{B}^{c}|\textbf{A}^{c})
\end{gather*}
\begin{gather*}
\textbf{P}(\textbf{A}^{c} \cap \textbf{B}^{c} ) \, = \, 0.4 \times 0.2 \, = \, 0.08
\end{gather*}
The probability of missing both buses is
\(0.08\) or
\(8\%\text{.}\)
\begin{gather*}
\end{gather*}
