To solve the problem, we start with the Poisson probability formula:

For a Poisson distribution with parameter $\lambda$, the probability of exactly $k$ successes is:
$$P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$$

Step 1: Set up the equation

We are told the probability of exactly 1 person passing equals the probability of exactly 2 people passing:
$$P(X=1) = P(X=2)$$

Step 2: Substitute the Poisson formula

$$\frac{\lambda^1 e^{-\lambda}}{1!} = \frac{\lambda^2 e^{-\lambda}}{2!}$$

Step 3: Simplify the equation

Cancel $e^{-\lambda}$ (non-zero) from both sides:
$$\frac{\lambda}{1} = \frac{\lambda^2}{2}$$

Multiply both sides by 2:
$$2\lambda = \lambda^2$$

Rearrange:
$$\lambda^2 - 2\lambda = 0 \implies \lambda(\lambda - 2) = 0$$

Step 4: Choose the valid solution

Since $\lambda > 0$, we get $\lambda = 2$.

Answer: $\boxed{2}$

作者声明:本文包含人工智能生成内容。