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Poisson: injection reaction with distribution

Short answer

The number of reactions is approximately Poisson with λ = 2000*0.001 = 2, so pmf is e^{-2}2^x / x!. Probability of exactly 3 is 0.1804, and more than 2 is 0.3233.

If the probability that an individual suffers a bad reaction from a certain injection is 0.001, among 2000 individual a) obtain probability distribution function for suffering bad reaction b) determine the probability that (i) exactly 3 individuals will suffer bad reaction (ii) more than 2, individuals will suffer bad reaction

n = 2000 and p = 0.001, so the number of bad reactions X is binomial. Since n is large and p is small, it is approximated by a Poisson distribution with

λ = np = 2000 × 0.001 = 2.

a) Probability distribution function

P(X = x) = e^(−λ) λˣ / x! = e⁻² 2ˣ / x!, x = 0, 1, 2, …

Here mean = variance = λ = 2 and e⁻² = 0.135335.

b) Probabilities

P(0) = e⁻²            = 0.1353
P(1) = e⁻² · 2        = 0.2707
P(2) = e⁻² · 2²/2!    = 0.2707
P(3) = e⁻² · 2³/3!    = 0.1804

(i) Exactly 3: P(X = 3) = 0.135335 × 8/6 = 0.1804

(ii) More than 2: P(X > 2) = 1 − [P(0) + P(1) + P(2)] = 1 − (0.1353 + 0.2707 + 0.2707) = 1 − 0.6767 = 0.3233

Answer: (i) 0.1804, (ii) 0.3233

✓ Answer checked by Claude Sonnet 5.5

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