为了减少道路上的超速事件数量，S国的交警在三个繁忙的交通路口 A、B 和 C 设置了交通摄像头，以监控通过这些路口的车辆速度。路口 A、B 和 C 的摄像头捕获的超速车辆平均数分别为每 3 小时 2 辆、每 4 小时 5 辆和每小时 $\lambda$ 辆。假设这三个路口的摄像头捕获的超速车辆数服从泊松分布。

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).