If the probability that an individual suffers a bad reaction from an injection of a given
 serum is 0.002, determine the probability that out of 400 individuals
(i) exactly 2
(ii) more than 3
(iii) at least one
 individuals suffer from bad reaction. 

To solve these problems numerically, we can use the binomial probability formula:

\[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k} \]

Where:
- \( P(X = k) \) is the probability of getting exactly \( k \) successes.
- \( n \) is the number of trials (individuals in this case).
- \( p \) is the probability of success (probability of suffering a bad reaction).
- \( \binom{n}{k} \) is the binomial coefficient, also known as "n choose k", calculated as \( \frac{n!}{k! \cdot (n - k)!} \).

Let's solve each part:

(i) Probability of exactly 2 individuals suffering from a bad reaction out of 400:
\[ P(X = 2) = \binom{400}{2} \cdot (0.002)^2 \cdot (1 - 0.002)^{400 - 2} \]

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