If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001. The probability that out of 2000 individuals, more than two individuals suffer from bad reaction is:
[Given that \(e^{-2} = 0.13534\)]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.3233
Here n = 2000 is large and p = 0.001 is small, so the number X of individuals with a bad reaction follows the Poisson distribution with
\(\lambda = np = 2000 \times 0.001 = 2\)
\( \displaystyle P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \)
“More than two” is the complement of “at most two”:
\(P(X \gt 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]\)
\( \displaystyle = 1 - e^{-2}\left(1 + 2 + \frac{2^2}{2!}\right) \)
\(= 1 - 5e^{-2}\)
\(= 1 - 5 \times 0.13534\)
\(= 1 - 0.6767 = 0.3233\)
Why the other options are incorrect
Option (1) 0.6767 — Incorrect. This is \(P(X \le 2)\), the probability of at most two, not more than two.
Option (3) 0.18044 — Incorrect. \(P(X = 3)\) alone is \(\frac{4}{3}e^{-2} \approx 0.1805\), but “more than two” also includes X = 4, 5, 6, …
Option (4) 0.9753 — Incorrect. The required probability is \(1 - 0.6767 = 0.3233\).