I have a doubt about the following problem:
the number of visits is an aleatory variable that has a Poisson distribution with a rate of 2 visits per minute. We know that in 4 minutes there were 5 visits. What is the probability that more visits occurred in the first 2 minutes than in the las 2 minutes?
My thoughts:
$\lambda=2$ and $\lambda t=2\times 2=4$. We find the probabilities for the cases where of 3,4,or 5 visits in the first 2 minutes:
$P(first2minutes)=P(3,4)+P(4,4)+P(5,4)=0.547$
where $P(x,\lambda t)=\frac{(\lambda t)^xe^{-\lambda t}}{x!}$
Is this correct?
Thanks.
Best Answer
Let $B$ be the event there were $5$ visits in $4$ minutes. Let $A$ be the event there were more visits in the first $2$ minutes than in the last $2$. We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$.
The number of visits in $4$ minutes has Poisson distribution parameter $8$. It follows that $\Pr(B)=e^{-8}\frac{8^5}{5!}$.
The event $A$ can happen in $3$ ways: (i) $3$ in the first $2$ minutes, and $2$ in the next $2$ minutes; (ii) $4$ and $1$; $5$ and $0$, The number of visits in $2$ minutes has Poisson distribution parameter $4$.
(i) The probability of $3$ and $2$ is $e^{-4}\frac{4^3}{3!}e^{-4}\frac{4^2}{2!}$. We can find similar expressions for the probabilities of (ii) and (iii). Add to find $\Pr(A\cap B)$.
Remark: There were some errors in the proposed solution. First, the problem is a conditional probability problem, and the computation did not take account of that. Second, the probabilities of $3$ and $2$, $4$ and $1$, and $5$ and $0$ were not calculated correctly, since the calculations did not take into account what happens in the second $2$ minute interval.