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Poisson: typist mistakes per page

Short answer

Using Poisson distribution with mean 2.6, probability of no mistake is 0.0743, at least two mistakes is 0.7326, and at most three mistakes is 0.7360.

A typist made 2.6 mistakes per page on average, find the probability that the page typed by him, i) there is no mistake ii) at least two mistakes iii) at most 3 mistakes.

Mistakes per page are rare, independent events over a page, so X ~ Poisson with mean λ = 2.6:

P(X = x) = e⁻²·⁶ (2.6)ˣ / x!, with e⁻²·⁶ = 0.074274.

x   P(X = x)
0   0.074274
1   0.074274 × 2.6        = 0.193111
2   0.193111 × 2.6/2      = 0.251045
3   0.251045 × 2.6/3      = 0.217572

i) No mistake

P(X = 0) = 0.0743

ii) At least two mistakes

P(X ≥ 2) = 1 − P(0) − P(1) = 1 − 0.0743 − 0.1931 = 0.7326

iii) At most three mistakes

P(X ≤ 3) = 0.0743 + 0.1931 + 0.2510 + 0.2176 = 0.7360

Answer: (i) 0.0743, (ii) 0.7326, (iii) 0.7360

✓ Answer checked by Claude Sonnet 5.5

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