The number of accidents on a certain road has a Poisson distribution with mean $$\(0.4\)$$ per 50-day period. Find the probability that there will be fewer than 3 accidents during a year ( 365 days).
Exam No：9709_m20_qp_62 Year：2020 Question No：4(a)
Answer:
\(\lambda(=0.4 \times 365 \div 50)=2.92\)
\(\mathrm{e}^{-2.92}\left(1+2.92+\frac{2.92^{2}}{2}\right)\)
\[
=0.441(3 \mathrm{sf})
\]
\(\mathrm{e}^{-2.92}\left(1+2.92+\frac{2.92^{2}}{2}\right)\)
\[
=0.441(3 \mathrm{sf})
\]
Knowledge points:
6.1.1 use formulae to calculate probabilities for the distribution $\text { Po }(\lambda)$
Solution:
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