The average number of railway accidents occurring along the North-South (NS) line in any one month is $m$. Accidents occur independently of one another and the probability of a railway accident occurring due to technical fault is given by $p$. The probability that there is at least one railway accident occurring along the NS line in any one week is known to be 0.25. A week is taken as a fourth part of a month.

1. Define the distribution for the number of accidents in a week.

Let $W$ be the number of railway accidents on the NS line in one week.
Since a week is a fourth part of a month, the average number of accidents in a week is $\frac{m}{4}$.
Given that accidents occur independently at a constant average rate, $W \sim \text{Po}\left(\frac{m}{4}\right)$.

2. Formulate an equation using the given probability of at least one accident.

We are given that $\text{P}(W \ge 1) = 0.25$.
Using the complement rule:
$1 - \text{P}(W = 0) = 0.25$
$1 - e^{-\frac{m}{4}} = 0.25$
$e^{-\frac{m}{4}} = 0.75$

3. Solve for the exact value of $m$.

Take the natural logarithm on both sides:
$-\frac{m}{4} = \ln(0.75)$
$m = -4\ln\left(\frac{3}{4}\right)$
$m = 4\ln\left(\frac{4}{3}\right)$

1. Find the average number of accidents in a week ($\lambda$) from the context of part (i).

Let $W$ be the number of railway accidents along the NS line in a week. Assuming accidents occur randomly and independently, $W$ follows a Poisson distribution, $W \sim \text{Po}(\lambda)$.
Given $P(W \ge 1) = 0.25$:
$1 - P(W = 0) = 0.25 \implies e^{-\lambda} = 0.75$
$\lambda = -\ln(0.75) = \ln\left(\frac{4}{3}\right)$

2. Calculate the probability that exactly 2 accidents occur along the NS line in a week.

$P(W = 2) = \frac{e^{-\lambda} \lambda^2}{2!}$
$P(W = 2) = \frac{0.75 \left(\ln\frac{4}{3}\right)^2}{2} = \frac{3}{8}\left(\ln\frac{4}{3}\right)^2$

3. Determine the conditional probability that exactly one of the two accidents is due to a technical fault.

Given that 2 accidents occurred, the number of accidents due to technical faults follows a Binomial distribution $B(2, p)$.
The probability that exactly one of the 2 accidents is due to a technical fault is:
$P(1 \text{ fault} \mid W=2) = \binom{2}{1} p^1 (1-p)^1 = 2p(1-p)$

4. Multiply the probabilities to find the overall required probability.

$\text{Required probability} = P(W = 2) \times P(1 \text{ fault} \mid W=2)$
$= \frac{3}{8}\left(\ln\frac{4}{3}\right)^2 \times 2p(1-p)$
$= \frac{3}{4}p(1-p)\left(\ln\frac{4}{3}\right)^2$

1. Define the distribution of accidents for both lines in one week.

Let $X$ be the number of accidents on the NS line and $Y$ be the number of accidents on the EW line in a week.
From part (i), the weekly average for the NS line is $\lambda_X = \frac{m}{4} = \ln\left(\frac{4}{3}\right)$.
For the EW line, the average per month is $2$, so the average per week is $\lambda_Y = \frac{2}{4} = 0.5$.
Assuming the accidents follow a Poisson distribution based on the constant average rate and independence, $X \sim \text{Po}\left(\ln\left(\frac{4}{3}\right)\right)$ and $Y \sim \text{Po}(0.5)$.

2. Set up the conditional probability expression.

We need to find $P(Y = 0 \mid X + Y = 3)$.
Using the formula for conditional probability:
$P(Y = 0 \mid X + Y = 3) = \frac{P(Y = 0 \text{ and } X + Y = 3)}{P(X + Y = 3)}$
Since $Y = 0$ and $X + Y = 3$ implies $X = 3$:
$P(Y = 0 \mid X + Y = 3) = \frac{P(Y = 0 \text{ and } X = 3)}{P(X + Y = 3)}$
Because accidents along the NS and EW lines are independent:
$P(Y = 0 \text{ and } X = 3) = P(Y = 0) P(X = 3)$

3. Calculate the required probability using the sum of independent Poisson distributions.

Since $X$ and $Y$ are independent, $X + Y \sim \text{Po}(\lambda_X + \lambda_Y)$, where $\lambda_X + \lambda_Y = \ln\left(\frac{4}{3}\right) + 0.5 \approx 0.78768$.

Calculating the numerator:
$P(Y = 0) P(X = 3) = \left( \frac{e^{-0.5} (0.5)^0}{0!} \right) \left( \frac{e^{-\ln(4/3)} (\ln(4/3))^3}{3!} \right) \approx 0.0018049$

Calculating the denominator:
$P(X + Y = 3) = \frac{e^{-0.78768} (0.78768)^3}{3!} \approx 0.037048$

Therefore,
$P(Y = 0 \mid X + Y = 3) = \frac{0.0018049}{0.037048} \approx 0.048718$

(Note: This simplifies algebraically to $\left(\frac{\lambda_X}{\lambda_X + \lambda_Y}\right)^3$.)

Rounding to 3 significant figures, the probability is $0.0487$.

1. Find the probability of at most 4 accidents occurring along the EW line in a single month.

Let $X$ be the number of railway accidents on the EW line in a month.
Since the average is 2, $X \sim \text{Po}(2)$.
$P(X \le 4) = e^{-2}\left(1 + 2 + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!}\right) = 7e^{-2} \approx 0.94735$

2. Define the binomial random variable for the 5-year period.

In 5 years, there are $5 \times 12 = 60$ months.
Let $Y$ be the number of months with at most 4 accidents.
$Y \sim \text{B}(60, 7e^{-2})$
We need to find $P(Y > 52)$.

3. Determine and state the suitable approximation.

We test the parameters for the approximation:
$E(Y) = np = 60 \times 7e^{-2} \approx 56.84$
$n(1-p) = 60(1 - 7e^{-2}) \approx 3.159$
Since $n = 60 \ge 50$ is large and $n(1-p) < 5$, the Normal approximation is not suitable. Instead, we use the Poisson approximation for the number of months with more than 4 accidents.
Let $W = 60 - Y$ be the number of months with more than 4 accidents.
$W \sim \text{B}(60, 1 - 7e^{-2})$
The suitable approximation is $W \sim \text{Po}(\lambda)$ where $\lambda = 60(1 - 7e^{-2}) \approx 3.1592$.

4. Calculate the final approximated probability.

$P(Y > 52) = P(60 - W > 52)$
$= P(W < 8)$
$= P(W \le 7)$
Using $W \sim \text{Po}(3.1592)$:
$P(W \le 7) \approx 0.98427 \approx 0.984$ (to 3 s.f.)