Understanding the Probability Problem

The question asks for the probability that out of 2000 individuals, more than two will suffer a bad reaction from an injection, given the probability of a bad reaction for a single individual is 0.001. This scenario involves a large number of trials (individuals) and a small probability of success (bad reaction) for each trial. This is a classic case where the Poisson distribution can be used as an approximation to the binomial distribution.

Identifying the Distribution and Parameters

We are dealing with a large number of independent trials (each individual's reaction is independent) and a small probability of a specific event (bad reaction) occurring in each trial. This fits the conditions for using the Poisson distribution.

The parameters are:

• Number of trials, \(n = 2000\)
• Probability of success (bad reaction) in a single trial, \(p = 0.001\)

For the Poisson distribution, the mean number of successes (\(\lambda\)) is calculated as \( \lambda = n \times p \).

Let's calculate the mean (\(\lambda\)):

\( \lambda = 2000 \times 0.001 = 2 \)

So, the average number of individuals expected to have a bad reaction is 2.

Calculating the Desired Probability

We want to find the probability that more than two individuals will suffer a bad reaction. In terms of the Poisson random variable \(X\) (the number of individuals with a bad reaction), we want to find \( P(X > 2) \).

The Poisson probability mass function is given by: \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \(k\) is the number of successes.

The probability \( P(X > 2) \) is equal to \( 1 - P(X \le 2) \).

\( P(X \le 2) \) is the probability that the number of individuals with a bad reaction is less than or equal to 2. This includes the cases where the number of bad reactions is exactly 0, 1, or 2.

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

Step-by-Step Probability Calculation

We need to calculate \( P(X=0) \), \( P(X=1) \), and \( P(X=2) \) using the Poisson formula with \( \lambda = 2 \).

1. Calculate \( P(X=0) \): \( P(X=0) = \frac{e^{-2} 2^0}{0!} = \frac{e^{-2} \times 1}{1} = e^{-2} \)
2. Calculate \( P(X=1) \): \( P(X=1) = \frac{e^{-2} 2^1}{1!} = \frac{e^{-2} \times 2}{1} = 2e^{-2} \)
3. Calculate \( P(X=2) \): \( P(X=2) = \frac{e^{-2} 2^2}{2!} = \frac{e^{-2} \times 4}{2} = 2e^{-2} \)

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

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

We need the value of \( e^{-2} \). Using a calculator, \( e^{-2} \approx 0.13534 \).

So, \( P(X \le 2) \approx 5 \times 0.13534 = 0.6767 \)

Finally, calculate \( P(X > 2) \):

\( P(X > 2) = 1 - P(X \le 2) \approx 1 - 0.6767 = 0.3233 \)

Comparing this result to the given options, the closest value is 0.32.

Summary of Probability Calculation

The probability that more than two individuals will suffer a bad reaction is approximately 0.32.

Revision Table: Probability Concepts

Additional Information: Poisson Distribution Conditions

The Poisson distribution is a suitable model for the number of events (like bad reactions) occurring within a fixed interval or sample size under certain conditions:

• The events are independent of each other. The reaction of one individual does not affect another.
• The average rate of events (\(\lambda\)) is constant. The probability of a bad reaction is the same for each individual.
• The events occur randomly in the interval.
• It is often used when the number of trials \(n\) is large and the probability of success \(p\) is small (typically \(n \ge 50\) and \(np < 5\) or \(p \le 0.1\)). In this case, \(n=2000\) (large) and \(p=0.001\) (small), and \(np = 2\). This confirms the appropriateness of using the Poisson approximation for this probability calculation.

Understanding when to apply the Poisson distribution is crucial for solving problems involving rare events in large populations or over long periods.