Poisson vs binomial mix-up

I have a problem where I am given that accidents occur at a Poisson rate of 3 per day. I am asked to find the probability that in January (31 days long), exactly 3 days will be accident-free. The correct way to do this is to find a Poisson rate $P(N(1)=0),$ so the probability of 0 accidents per 1 day using $$P(N(1)=0) = \frac{(3\cdot1)^0 e^{-3}}{0!} = 0.0498$$ (since $\lambda = 3$), and then do $$\binom{31}{3} (0.0498)^3 (1-0.0498)^{28} = 0.132759$$ Intuitively, this answer ring the bell to me, but my initial approach was the following: find $P(N(3)=0),$ or a Poisson process estimation of probability that in 3 days, 0 accidents happen, and count $$\binom{31}{3} P(N(3) = 0) = \binom{31}{3} \frac{(3\cdot3)^{0} e^{-9}}{0!} = 0.555.$$ This is clearly way off, and I would be happy if someone could explain me what is it that I found in the second case and why it makes no sense.

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You can interpret $\mathbb P(N(3) = 0)$ as the probability that over the course of three consecutive days, no accidents happen. Clearly (a) that is not what you are interested in and (b) there are way less than ${31 \choose 3}$ of them. :) –  cardinal Oct 13 '11 at 0:53
Note that in your second answer, you have forgotten to multiply by $(1-e^-3)^{28}$, the probability of at least one accident on the remaining $28$ days, Indeed, your first (correct) answer is $\binom{31}{3}(e^{-3})^3(1-e^{-3})^{28} = \binom{31}{3}(e^{-9})(1-e^{-3})^{28}$. I don't quite agree with @cardinal that $P(N(3) = 0)$ is the probability of no accidents on three consecutive days; $P(N(3) = 0)$ is actually the probability of no accidents on three given days (which don't necessarily have to be consecutive days). –  Dilip Sarwate Oct 13 '11 at 1:23
@DilipSarwate: Note that I said "You can interpret...", the intent being to provide intuition. But, I think your comment gets to the heart of the matter. I tend to be more obtuse/indirect with mine... –  cardinal Oct 13 '11 at 1:27
@cardinal Apologies for misreading your remark as saying that probability of no accidents over three consecutive days was the sole interpretation of $P(N(3) = 0)$. Probably "(b) there are way less than $\binom{31}{3}$ of them" compounded the wrong impression that I got since with the three given days interpretation, $\binom{31}{3}$ is the right number to use to get the probability the OP sought. –  Dilip Sarwate Oct 13 '11 at 1:43
@DilipSarwate: Yes, I can see how my second remark would compound the problem, when read that way. I fear I was a bit too flippant on both counts. :) –  cardinal Oct 13 '11 at 1:47