Probability - Poisson arrival of rain I'm trying to solve this Poisson problem. A rain shower lasts 10 minutes and in that time deposits $10^6$ raindrops over 100 $m^2$. a) What is the probability of at least one drop landing in 1 $cm^2$ b)On average, how much time elapses before a raindrop hits the 1 $cm^2$ area. I understand the first part, the intensity is 1 drop/$cm^2$ so the probability is of at least one is $P(x\ge1)=1-\frac{(1^0)e^{-1}}{0!}=0.632$ But I'm not sure about the second part. I don't see how the number of drops, the time interval, and the two areas relate to each other to appear in a Poisson proccess. Particularly since I'm apparently not trying to find the probability of no drops after some amount of time.

There are $100(100^2) = 10^6$ cm$^2$ in the full area of $100$ m$^2$. Therefore, in $10$ minutes, an average of one raindrop fell in each cm$^2$, and the rate is $0.1$ raindrop/cm$^2$/minute.

You answer to part (a) is correct: The probability that at least one raindrop falls in any given cm$^2$ is

$$ P(X >= 1) = 1-P(0) = 1-\frac{(10 \cdot 0.1)^0}{0!}e^{-10 \cdot 0.1} = 1-e^{-1} \doteq 0.63212 $$

For part (b), since the rate is $\lambda = 0.1$ raindrop/cm$^2$/minute, the expected time is $1/\lambda = 10$ minutes$\cdot$cm$^2$/raindrop. As we are concerned with only $1$ cm$^2$, the expected time is $10$ minutes/raindrop.