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?
Best Answer
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 $\lambda$.