The monthly expectation of a two-wheeler accident at a particular railway crossing is 0.02. What is the probability that in a specific month, no accident occurred ? (given \( e^{-0.02}=0.9802 \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.9802
Accidents are rare events counted over a fixed period, so the number of accidents X in a month follows a Poisson distribution with mean \( \lambda=0.02 \).
\( \displaystyle P(X=r)=\frac{e^{-\lambda}\lambda^r}{r!} \)
\( \displaystyle P(X=0)=\frac{e^{-0.02}(0.02)^0}{0!} \)
\( \displaystyle =e^{-0.02} \)
\( \displaystyle =0.9802 \)
Why the other options are incorrect
Option (1) 0.8904 — Incorrect. P(X = 0) is \( e^{-0.02}=0.9802 \), not 0.8904.
Option (2) 0.7803 — Incorrect. P(X = 0) is \( e^{-0.02}=0.9802 \), not 0.7803.
Option (4) 0.0198 — Incorrect. This is \( 1-e^{-0.02} \), the probability of at least one accident.