Problem Analysis: Injection Reaction Probability

The question asks for the probability of observing more than two adverse reactions ('bad reaction') in a group of 2,000 individuals, given a small individual probability of a bad reaction.

• Number of individuals (trials), $n = 2000$.
• Probability of a bad reaction for one individual, $p = 0.001$.
• We need to find the probability of more than two bad reactions, $P(X > 2)$.

Since $n$ is large and $p$ is small, this scenario can be effectively modeled using the Poisson distribution as an approximation to the binomial distribution.

Poisson Approximation Calculation

The parameter $\lambda$ (lambda) for the Poisson distribution is calculated as:

$ \lambda = n \times p $

Substituting the values:

$ \lambda = 2000 \times 0.001 = 2 $

The Poisson probability mass function is given by:

$ P(X=k) = \frac{e^{-\lambda} \lambda^{k}}{k!} $

We want to calculate $P(X > 2)$. It's easier to calculate the complement probability, $P(X \le 2)$, and subtract it from 1:

$ P(X > 2) = 1 - P(X \le 2) $

$ P(X \le 2) = P(X=0) + P(X=1) + P(X=2) $

Calculating Individual Probabilities ($\lambda=2$)

1. Probability of zero bad reactions ($k=0$):
   $ P(X=0) = \frac{e^{-2} 2^{0}}{0!} = \frac{e^{-2} \times 1}{1} = e^{-2} $
2. Probability of one bad reaction ($k=1$):
   $ P(X=1) = \frac{e^{-2} 2^{1}}{1!} = \frac{e^{-2} \times 2}{1} = 2e^{-2} $
3. Probability of two bad reactions ($k=2$):
   $ P(X=2) = \frac{e^{-2} 2^{2}}{2!} = \frac{e^{-2} \times 4}{2} = 2e^{-2} $

Final Probability Calculation

Now, sum these probabilities to find $P(X \le 2)$:

$ P(X \le 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2} $

Finally, calculate the desired probability $P(X > 2)$:

$ P(X > 2) = 1 - P(X \le 2) = 1 - 5e^{-2} $

This can also be written as:

$ P(X > 2) = 1 - \frac{5}{e^{2}} $

This result matches Option A.