1. **State the problem:** Given the probability of a bad reaction from serum injection is 0.001, and 2000 individuals are injected, find: (i) The probability exactly 3 individuals suffer a bad reaction. (ii) The probability more than 2 individuals suffer a bad reaction. 2. **Identify distribution:** The number of bad reactions, $X$, follows a Binomial distribution with parameters $n=2000$, $p=0.001$. Since $n$ is large and $p$ is small, approximate $X$ by a Poisson distribution with parameter $\lambda=np=2000 \times 0.001=2$. 3. **Solve (i): Probability exactly 3 bad reactions:** The Poisson probability mass function is: $$P(X=k) = \frac{e^{-\lambda} \lambda^{k}}{k!}$$ For $k=3$: $$P(X=3) = \frac{e^{-2} 2^{3}}{3!} = \frac{e^{-2} \times 8}{6} = \frac{4}{3} e^{-2}$$ 4. **Solve (ii): Probability more than 2 bad reactions:** $$P(X>2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$ Calculate each term: $$P(X=0) = \frac{e^{-2} 2^{0}}{0!} = e^{-2}$$ $$P(X=1) = \frac{e^{-2} 2^{1}}{1!} = 2 e^{-2}$$ $$P(X=2) = \frac{e^{-2} 2^{2}}{2!} = 2 e^{-2}$$ Summing: $$P(X \leq 2) = e^{-2} + 2 e^{-2} + 2 e^{-2} = 5 e^{-2}$$ Hence: $$P(X>2) = 1 - 5 e^{-2}$$ **Final answers:** (i) $P(X=3) = \frac{4}{3} e^{-2}$ (ii) $P(X>2) = 1 - 5 e^{-2}$