Problem source

I can choose one of three routes to cycle to school. Via Angle Avenue the distance is 5$\,$km, and I am held up at a level crossing for $A$ minutes, where $A$ is a continuous random variable uniformly distributed between $0$ and 10. Via Bend Boulevard the distance is 4$\,$km, and I am delayed, by talking to each of $B$ friends for 3$\,$minutes, for a total of $3B$ minutes, where $B$ is a random variable whose distribution is Poisson with mean 4. Via Detour Drive the distance should be only 2$\,$km, but in addition, due to never-ending road works, there are five places at each of which, with probability $\frac{4}{5},$ I have to make a detour that increases the distance by 1$\,$km. Except when delayed by talking to friends or at the level crossing, I cycle at a steady 12$\,$km$\,$h$^{-1}$. For each of the three routs, calculate the probability that a journey lasts at least 27 minutes. 
Each day I choose one of the three routes at random, and I am equally likely to choose any of the three alternatives. One day I arrive at school after a journey of at least 27 minutes. What is the probability that I came via Bend Boulevard?
Which route should I use all the time: 
\begin{questionparts}
\item if I wish my average journey time to be as small as possible; 
\item if I wish my journey time to be less than 32 minutes as often as possible?
\end{questionpart}
Justify your answers.

Solution source

$A \sim 5\cdot 5 + U[0,10]$
$B \sim 4 \cdot 5 + 3 \textrm{Po}(4)$
$C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5$

\begin{align*}
&& \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\
&& \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\
&&&= \mathbb{P}(Po(4) \leq 2) \\
&&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\
&&&= 0.23810\ldots \\
&& \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\
&&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\
&&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\
&&&= 0.26272
\end{align*}

\begin{align*}
\mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\
&= \frac{\frac13 \cdot  0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot  0.23810\ldots  + \frac13 \cdot 0.26272} \\
&= 0.3397\ldots \\
&= 0.340 \, \, (3\text{ s.f.})
\end{align*}

\begin{questionparts}
\item \begin{align*}
\mathbb{E}(A) &= 25 + 5 &= 30 \\
\mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\
\mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30
\end{align*}

$A$ and $C$ are equally good.
\item \begin{align*}
\mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\
\mathbb{P}(B \leq 32) &=  \mathbb{P}(Po(4) \leq 4)  \\
&= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\
\mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\
&= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\
&= 1 - \frac{4^5}{5^5} &=0.67232
\end{align*}

So you should choose route $A$.
\end{questionparts}