Problem source

\textit{Tabulated values of }${\Phi}(\cdot)$\textit{, the cumulative distribution function of a standard normal variable, should not be used in this question.} 
 Henry the  commuter lives in Cambridge and his working day starts at his office in London at 0900. He catches the 0715 train to King's Cross with probability $p$, or the 0720 to Liverpool Street with probability $1-p$. Measured in minutes, journey times for the first train are $N(55,25)$ and for the second are $N(65,16)$. Journey times from King's Cross and Liverpool Street to his office are $N(30,144)$ and $N(25,9)$, respectively. Show that Henry is more likely to be late for work if he catches the first train.
Henry makes $M$ journeys, where $M$ is large. Writing $A$ for $1-{\Phi}(20/13)$  and $B$ for $1-{\Phi}(2)$, find, in terms of $A$, $B$, $M$ and $p$, the expected number, $L$, of times that Henry will be late and show that for all possible values of $p$,
$$BM \le L \le AM.$$ 
Henry noted that in 3/5 of the occasions when he was late, he had caught the King's Cross train. Obtain an estimate of $p$ in terms of $A$ and $B$.
[A random variable is said to be $N\left({{\mu}, {\sigma}^2}\right)$ if it has a normal distribution with mean ${\mu}$ and variance ${\sigma}^2$.]

Solution source

If Henry catches the first train, his journey time is $N(55+30,25+144) = N(85,13^2)$. He is on time if the journey takes less than $105$ minutes, $\frac{20}{13}$ std above the mean.

If he catches the second train, his journey times is $N(65+25, 16+9) = N(90, 5^2)$. He is on time if his journey takes less than $80$ minutes, ie $\frac{10}{5} = 2$ standard deviations above the mean. This is more likely than from the first train.

$A = 1 - \Phi(20/13)$ is the probability he is late from the first train.
$B = 1 - \Phi(2)$ is the probability he is late from the second train.

The expected number of lates is $L = M \cdot p \cdot A + M \cdot (1-p) \cdot B$, since $B \leq A$ we must have $BM \leq L \leq AM$

\begin{align*}
&& \frac35 &= \frac{pA}{pA + (1-p)B} \\
\Rightarrow && 3(1-p)B &= 2pA \\
\Rightarrow && p(2A+3B) &= 3B \\
\Rightarrow && p &= \frac{3B}{2A+3B}
\end{align*}