The number of accidents in a day at a busy crossing follows a Poisson distribution with mean 0.4. The probability that there is no accident at the crossing on a given day is (Given \(e^{-0.4} = 0.6703\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.6703
Let X be the number of accidents in a day. X follows a Poisson distribution with \(m = 0.4\):
\( \displaystyle P(X = r) = \frac{e^{-m}\,m^{r}}{r!} \)
No accident means r = 0:
\( \displaystyle P(X = 0) = \frac{e^{-0.4}\,(0.4)^{0}}{0!} \)
\(= e^{-0.4} = 0.6703\)
Why the other options are incorrect
Option (2) 0.3297 — Incorrect. This is 1 − 0.6703, the probability of at least one accident.
Option (3) 0.2681 — Incorrect. This is \(P(X = 1) = 0.4 \times 0.6703\), exactly one accident.
Option (4) 0.4 — Incorrect. 0.4 is the mean number of accidents in a day, not a probability of no accident.