Question

If the probability of a bad reaction from a certain injection is \(0.001\), then the chance that out of 2000 individuals more than two will get a bad reaction is given by:

• A. \(1 - \frac{5}{e^2}\)
• B. \(\frac{5}{e^2}\)
• C. \(1 - \frac{2}{e^2}\)
• D. \(\frac{4}{e^2}\)

Hint:

In Poisson approximation, always compute \(\lambda = np\) first. For "more than" type questions, using complement \(1 - P(X \leq k)\) saves significant time and effort.

Answer 1

The Correct Option is A

Concept: This problem is a classical application of Poisson approximation to the Binomial distribution, which is used when:
• The number of trials \(n\) is very large,
• The probability of success \(p\) is very small,
• The product \(np = \lambda\) is finite and moderate. Instead of directly computing binomial probabilities (which would involve very large factorials and cumbersome calculations), we simplify the problem using the Poisson distribution: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] This significantly reduces computational complexity while giving a very accurate approximation.

Step 1: Carefully identify all given quantities.
We are given: \[ n = 2000 \quad \text{(number of individuals)} \] \[ p = 0.001 \quad \text{(probability of bad reaction)} \] Now compute: \[ \lambda = np = 2000 \times 0.001 = 2 \] Thus, the corresponding Poisson distribution has parameter: \[ \lambda = 2 \]

Step 2: Clearly interpret the requirement.
We need to find: \[ P(X > 2) \] This means the probability that more than 2 individuals out of 2000 experience a bad reaction.

Step 3: Use complement rule to simplify the problem.
Direct computation of \(P(X > 2)\) would require summing infinite terms. Instead, we use: \[ P(X > 2) = 1 - P(X \leq 2) \] This converts the problem into computing: \[ P(X \leq 2) = P(0) + P(1) + P(2) \]

Step 4: Compute each probability term in detail.
For \(X = 0\): \[ P(0) = \frac{e^{-2} \cdot 2^0}{0!} \] Since \(2^0 = 1\) and \(0! = 1\), we get: \[ P(0) = e^{-2} \] For \(X = 1\): \[ P(1) = \frac{e^{-2} \cdot 2^1}{1!} \] \[ P(1) = \frac{e^{-2} \cdot 2}{1} = 2e^{-2} \] For \(X = 2\): \[ P(2) = \frac{e^{-2} \cdot 2^2}{2!} \] \[ P(2) = \frac{e^{-2} \cdot 4}{2} = 2e^{-2} \]

Step 5: Add all probabilities carefully.
\[ P(X \leq 2) = e^{-2} + 2e^{-2} + 2e^{-2} \] Factor out \(e^{-2}\): \[ P(X \leq 2) = e^{-2}(1 + 2 + 2) \] \[ P(X \leq 2) = 5e^{-2} \]

Step 6: Apply complement to get final result.
\[ P(X > 2) = 1 - 5e^{-2} \] \[ P(X > 2) = 1 - \frac{5}{e^2} \] Final Answer: \[ \boxed{1 - \frac{5}{e^2}} \]