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Poisson: generators failing in warranty

Short answer

Poisson approximation with λ=np=3.2. P(exactly 3 fail)=0.2226; P(between 2 and 6 inclusive)=0.7842.

A heavy machinery manufacturer has 3840 large generators in the field that are under warranty. If the probability is 1/1200 that any one will fail during the given year, find the probability: i) That exactly 3 generators will fail during the given year? ii) That between 2 and 6 are fail during the given year?

n = 3840 generators, p = 1/1200 (small), so the number of failures X is approximately Poisson with

λ = np = 3840 / 1200 = 3.2, P(X = x) = e⁻³·² (3.2)ˣ / x!, e⁻³·² = 0.040762.

x   P(X = x)
2   0.040762 × 3.2²/2    = 0.208702
3   0.040762 × 3.2³/6    = 0.222616
4   0.222616 × 3.2/4     = 0.178093
5   0.178093 × 3.2/5     = 0.113979
6   0.113979 × 3.2/6     = 0.060789

i) Exactly 3 fail

P(X = 3) = 0.2226

ii) Between 2 and 6 fail

Taking the limits as included (2 ≤ X ≤ 6):

P = 0.2087 + 0.2226 + 0.1781 + 0.1140 + 0.0608 = 0.7842

(If "between" is read as strictly between, 3 ≤ X ≤ 5, the answer is 0.2226 + 0.1781 + 0.1140 = 0.5147.)

Answer: (i) 0.2226, (ii) 0.7842

✓ Answer checked by Claude Sonnet 5.5

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