To reduce the number of speeding incidents on the road, traffic police in Country S set up traffic cameras at 3 busy traffic Junctions A, B and C to monitor the speeds of vehicles passing through these junctions. The average number of speeding vehicles caught by the camera at Junctions A, B and C are 2 in every 3 hours, 5 in every 4 hours and $\lambda$ in every hour respectively. It is assumed that the number of speeding vehicles caught by the cameras at the three junctions followed Poisson distributions.

1. Define the combined random variable for the number of vehicles caught at Junctions A and B in an hour.

Let $X$ and $Y$ be the number of speeding vehicles caught at Junctions A and B in an hour respectively.
$X \sim \text{Po}(\frac{2}{3})$
$Y \sim \text{Po}(\frac{5}{4})$
Assuming the number of vehicles caught at each junction are independent, let $W = X + Y$ be the total number of vehicles caught at both junctions.
$W \sim \text{Po}(\frac{2}{3} + \frac{5}{4})$
$W \sim \text{Po}(\frac{23}{12})$

2. Calculate the probability of at least 2 speeding vehicles being caught.

$\text{P}(W \ge 2) = 1 - \text{P}(W \le 1)$
$\text{P}(W \ge 2) = 1 - \left( \text{P}(W = 0) + \text{P}(W = 1) \right)$
$\text{P}(W \ge 2) = 1 - e^{-\frac{23}{12}} \left( 1 + \frac{23}{12} \right)$
$\text{P}(W \ge 2) \approx 1 - 0.42903 \approx 0.571 \text{ (to 3 s.f.)}$

1. Define the Poisson parameters for each junction and the combined parameter for Junctions A and B.

Let $X_A$, $X_B$, and $X_C$ be the number of speeding vehicles caught in an hour at Junctions A, B, and C respectively.
$X_A \sim \text{Po}\left(\frac{2}{3}\right)$
$X_B \sim \text{Po}\left(\frac{5}{4}\right)$
$X_C \sim \text{Po}(\lambda)$
Let $W = X_A + X_B$ be the number of speeding vehicles caught at Junctions A and B in an hour.
Assuming independence, the sum follows a Poisson distribution:
$W \sim \text{Po}\left(\frac{2}{3} + \frac{5}{4}\right) \Rightarrow W \sim \text{Po}\left(\frac{23}{12}\right)$

2. Formulate the required conditional probability using the complement rule.

Let $Y = W + X_C$ be the total vehicles caught at all three junctions in an hour, so $Y \sim \text{Po}\left(\frac{23}{12} + \lambda\right)$.
We need to find $P(X_C \ge 1 \mid Y = 2)$.
Using the complement rule:
$P(X_C \ge 1 \mid Y = 2) = 1 - P(X_C = 0 \mid Y = 2)$
By the definition of conditional probability:
$P(X_C = 0 \mid Y = 2) = \frac{P(X_C = 0 \cap Y = 2)}{P(Y = 2)} = \frac{P(X_C = 0 \cap W = 2)}{P(Y = 2)}$

3. Calculate the constituent probabilities and simplify the final expression.

Since $X_C$ and $W$ are independent, $P(X_C = 0 \cap W = 2) = P(X_C = 0)P(W = 2)$:
$P(X_C = 0) = \text{e}^{-\lambda}$
$P(W = 2) = \text{e}^{-\frac{23}{12}} \frac{\left(\frac{23}{12}\right)^2}{2!}$
$P(Y = 2) = \text{e}^{-\left(\frac{23}{12} + \lambda\right)} \frac{\left(\frac{23}{12} + \lambda\right)^2}{2!}$
Substituting these into the conditional probability ratio:
$\begin{aligned} P(X_C = 0 \mid Y = 2) &= \frac{\text{e}^{-\lambda} \cdot \text{e}^{-\frac{23}{12}} \frac{\left(\frac{23}{12}\right)^2}{2}}{\text{e}^{-\left(\frac{23}{12} + \lambda\right)} \frac{\left(\frac{23}{12} + \lambda\right)^2}{2}} \\ &= \frac{\left(\frac{23}{12}\right)^2}{\left(\frac{23}{12} + \lambda\right)^2} \\ &= \left( \frac{\frac{23}{12}}{\frac{23 + 12\lambda}{12}} \right)^2 \\ &= \left( \frac{23}{12\lambda + 23} \right)^2 \end{aligned}$
Therefore, the required probability is:
$P(X_C \ge 1 \mid Y = 2) = 1 - \left( \frac{23}{12\lambda + 23} \right)^2$

1. Determine the distribution of the number of speeding vehicles caught at Junctions A and B in a day.

Let $A$ and $B$ be the number of speeding vehicles caught in a 24-hour day at Junction A and Junction B respectively.
For Junction A: $\lambda_A = 24 \times \frac{2}{3} = 16$
For Junction B: $\lambda_B = 24 \times \frac{5}{4} = 30$
Thus, $A \sim \text{Po}(16)$ and $B \sim \text{Po}(30)$.

2. State the normal approximations for $A$ and $B$.

Since $\lambda_A = 16 > 10$ and $\lambda_B = 30 > 10$, the Poisson distributions can be approximated by Normal distributions:
$A \sim \text{N}(16, 16)$ approximately.
$B \sim \text{N}(30, 30)$ approximately.

3. Find the distribution of the difference $A - B$.

Let $W = A - B$.
Assuming $A$ and $B$ are independent,
$\text{E}(W) = \text{E}(A) - \text{E}(B) = 16 - 30 = -14$
$\text{Var}(W) = \text{Var}(A) + \text{Var}(B) = 16 + 30 = 46$
Therefore, $W \sim \text{N}(-14, 46)$ approximately.

4. Apply a continuity correction to find the required probability.

We require $\text{P}(A > B)$, which is equivalent to $\text{P}(A - B > 0)$.
Since $A - B$ takes integer values, $\text{P}(A - B > 0) = \text{P}(A - B \ge 1)$.
Applying the continuity correction for the normal approximation, we calculate $\text{P}(W > 0.5)$.
Using a graphing calculator:
$\text{P}(W > 0.5) \approx 0.016263 \approx 0.0163$ (to 3 s.f.)

5. State the necessary assumption for the calculation to be valid.

Assumption: The number of speeding vehicles caught at Junction A and the number caught at Junction B are independent of each other (this is required to sum their variances).