I need help with this probability problem.

The number of fatal car accidents that happen in a specific region follows the Poisson distribution with a rate of 0.5 fatal car accidents per day

(a) What is the probability that a fatal car accident will happen in the next 2 days and after that accident within no more than one day, the next fatal car accident occurs?

(b) Calculate (approximately) the probability to wait more than 60 days for the occurrence of the 40th fatal car accident.

My solution is the following.

(a) Firstly I calculate the probability that a fatal car accident will happen in 2 days.

$0.5$ car accidents happen in $1$ day, so $1$ accident happens in $2$ days. That means that my new rate is $λ=1$. Considering the random variable

$$X=\text{number of car accidents}$$

we have that $X$ follows $\operatorname{Poisson}(λ=1)$. As a result, $P(X=1)=0.3679$.

Next, I consider a random variable $Z=\text{number of days until the next car accident}.$ Then, Z follows $\operatorname{Geo}(p=0.3679)$ (I am not sure about the $p$). As a result, $P(Z\le1)=0.3679$.

To calculate the asked probability I simply multiply the above probabilities: $P(X=1)\cdot P(Z\le1)$.

(b) I consider the random variable $Y=\text{number of days till the 40th car accident}$ Then $Y$ follows $\operatorname{Negative Binomial}(k=40,p=0.3679)$ and the asked probability is $P(Y>60)$.

Is my solution right? What is your opinion?

  • $\begingroup$ NB in public health policy and underwriting, these are not called "accidents" or at least they shouldn't be. Rather, they are collisions. $\endgroup$
    – AdamO
    Dec 5, 2018 at 17:53
  • $\begingroup$ Thanks for your remark. Actually I am not an English native speaker. $\endgroup$
    – George
    Dec 5, 2018 at 20:45

1 Answer 1


Inter-arrival times between independent Poisson events are exponentially distributed, with $\lambda=0.5$ (in days), i.e. $\mu=\frac{1}{\lambda}=2$. Let $T_1$ be the time (in days) of the first fatal accident, and $T_2$ be the extra time needed for the next fatal accident. In (a), we need to calculate $P(T_1<2, T_2<1)$ $=P(T_1<2)$ $ P(T_2<1)$. Each $T_i$ is exponentially distributed. You can substitute this: $P(T<t)=1-e^{\lambda t}$. Be careful while changing your $\lambda$.

  • $\begingroup$ I agree with your solution. Is $λ=2$ in both $T_1$ and $T_2$? $\endgroup$
    – George
    Dec 6, 2018 at 9:05
  • $\begingroup$ Yes, because you have a stationary process. $\endgroup$
    – gunes
    Dec 6, 2018 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.