Skip to main content

Poisson: accidents among 1000 taxi drivers

Short answer

For Poisson with mean 3, use P(0)=e^-3 and complement. About 1000*0.0498 gives approximately 50 drivers with no accidents; about 1000*0.3528 gives approximately 353 drivers with more than 3 accidents.

The number of accident in a year attributes to taxi drivers in a city follows Poisson distribution with mean 3. Out of 1000 taxi driver, find the approximately the number of driver with: i) No accidents in a year ii) More than 3 accident in a year

The number of accidents per driver per year X ~ Poisson with λ = 3: P(X = x) = e⁻³ 3ˣ / x!, with e⁻³ = 0.049787.

P(0) = 0.049787
P(1) = 0.149361
P(2) = 0.224042
P(3) = 0.224042

i) No accidents

P(X = 0) = 0.0498, so the expected number of drivers = 1000 × 0.0498 ≈ 50 drivers

ii) More than 3 accidents

P(X > 3) = 1 − [P(0) + P(1) + P(2) + P(3)] = 1 − (0.049787 + 0.149361 + 0.224042 + 0.224042) = 1 − 0.647232 = 0.3528

Expected number of drivers = 1000 × 0.3528 ≈ 353 drivers

Answer: (i) about 50 drivers, (ii) about 353 drivers

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 2 questions →