# Poisson distribution question, reading from the table

The number of cars passing over a certain bridge between 11pm and midnight has a Poisson density with lambda = 4.

In what proportion of nights would you expect more than one car to pass over the bridge during this time?

I've already worked out, reading off values from the poisson table http://mat.iitm.ac.in/home/vetri/public_html/statistics/poisson.pdf

1 car would be 0.0916
2 car would be 0.2381
3 car would be 0.4335
4 car would be 0.6288

How would you work out the value for "at least 1"? I thought that it might just be the sum of all these values but then I realised all these values are cumulative, so would it be:
The value for 4 - The value for 1?
0.6288 - 0.0916?

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Why don't you count 5 cars? And 6? 7,8, etc... all are more than once. – Raskolnikov Sep 14 '11 at 10:43
hmmm good question, I'm not sure actually, I just assumed we go from 1 to 4 because lambda is 4, but I suppose this may not be the case. But if there was no upper bound then we would need to know the probability of there being at least 1 car, plus 2 car, all the way up to infinity, wouldn't the probability also approach infinity? I'm confused.. – Arvin Sep 14 '11 at 10:57
No, the probability for 5 cars is smaller than the probability for 4. And the probability for 6 is smaller than both, and so on. The probabilities fall off exponentially fast, so that the total sum is finite. – Raskolnikov Sep 14 '11 at 11:00
Yes, but be mindful of two things. (1) If lambda = 4, "no car" has probability exp(-4) = 0.0183 (and not 0.0111 which is exp(-4.5)). (2) If "more than one car" means "at least two cars", you want the probability of "no car or one car", which is exp(-4)*(1+4) = 0.0916. But if you manage to sort out these two points... bingo. – Did Sep 14 '11 at 11:38
Seems your answer should be 1-0.0916=0.9084 then. – Did Sep 14 '11 at 16:05

However, rather than being probabilities of a given number of cars crossing the bridge during the hour, the four numbers you list actually are the cumulative probabilities of given numbers of cars crossing it. (As shown in wikipedia the Poisson probability mass function is given by $f(k;\lambda) = \lambda^k e^{-\lambda}/k!$ and the numbers you tabulated are $f(0;4)+f(1;4), f(0;4)+f(1;4)+f(2;4),$ ...) So, for the given numbers, it's better to refer to "at least one car", "at least two cars", etc.
The probability that it's at least one is $1$ minus the probability that it's exactly $0$.