Car accidents and the Poisson distribution
Answer
Let $X$ be the number of accidents per week. Then $X$ has the Poisson distribution
\[p(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}, x=0,1,2, \dots .\]
Since the average number of accidents happen per week is $5$, $\lambda=5$. Hence
\[p(x)=\frac{5^{x}e^{-5}}{x!}, x=0,1,2, \dots .\]
The probabality of no accident is
\[P[X=0]=\frac{5^{0}e^{-5}}{0!}=\frac{e^{-5}}{1}=e^{-5}=\frac{1}{e^{5}}=0.0067=6.7\%.\]
Note that $0! =1$.

4.8K
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 2423 views
- $5.00
Related Questions
- Confidence Interval,Standard Deviation,Mean
- What is the probability that the last person to board an airplane gets to sit in their assigned seat?
- Pulling balls out of a bin
- Stats project
- Product of Numbers from a Log Normal Distribution
- Probability of having a disease given a series of test results
- Help with probability proofs and matrices proofs (5 problems)
- Algebra Word Problem 3