Strange Poisson process problem without having the mean parameter

Question

I found a problem of probability which asks:

The arrival of clients to a bank can be considered as a Poisson process with parameter L. It is given that 4 clients arrived in the last hour. Considering this, find the probability that 2 of the clients arrived in the first 30 minutes and the other 2 in the last 30 minutes of that hour.

The supposed correct answer is given (p = 0,375) but I was trying to solve the problem, and I think that in the original statement is missing L. What I tried to do is to apply the definition of conditional probability. Let:

$$P(N(t)=n) = \frac{(Lt)^n}{n!} e^{-Lt}$$

be the probability that n clients arrive in a time t. Then, I consider that the time starts to count since the first client came, so the probability I should be looking for is:

$$ \frac{P(N(1/2)=1)}{P(N(1)=3)} = \frac{(L/2) e^{-L/2}}{\frac{(L)^3}{3!} e^{-L}} $$

I can't see how to simplify L. Am I doing wrong something or there is missing information?

Answer 1

Let $\lambda$ be the expected number of clients in a half-hour period.

The probability that $2$ arrive in the first half-hour is $$\frac{e^{-\lambda} \lambda^2}{2!}$$ and likewise for the second half-hour.

The probability that $4$ arrive in a one-hour period is $$\frac{e^{2\lambda}(2\lambda)^4}{4!}$$

So the conditional probability you require is $$\left(\frac{e^{-\lambda} \lambda^2}{2!}\right)^2\div \left(\frac{e^{2\lambda}(2\lambda)^4}{4!}\right)$$ $$=\frac38$$

All the $\lambda$ terms cancel.

Answer 2

Since ‘each 30 minute period is the same as any other, as far as Poisson knows’, consider that the probability that each client arrived in the first 30 minutes is p=0,5. The correct answer is found if we apply the Binomial distribution:

$$ Combinatory(4,2) \cdot 0.5^{2} \cdot 0.5² = 0.375$$

Thanks @lulu! Love you!