Poisson distribution problem about car accidents I have this Poisson distribution question, which I find slightly tricky, and I'll explain why.
The number of car accidents in a city has a Poisson distribution. In March the number was $150$, in April $120$, in May $110$ and in June $120$. Eight days are being chosen by random, not necessarily in the same month. What is the probability that the total number of accidents in the eight months will be $30$?
What I thought to do, is to say that during this period, the average number of accidents is $125$ a month, and therefore this is my $\lambda$. Then I wanted to go from a monthly rate to a daily rate, and here comes the trick. How many days are in a month? So I choose $30$, and then the daily rate is $\frac{100}{3}$, and so the required probability is $0.06$. Am I making sense, or am I way off the direction in this one? Thank you! 

One could say that $500$ accidents in $122$ days means the daily average is $500/122\approx 4.09836$, so that the average for eight days is
$$
8\times\frac{500}{122} \approx 32.78689,
$$
and then treat it as a Poisson-distribution problem with that expected value.  This yields
$$
\frac{e^{-32.78689} 32.78689^{30}}{30!} \approx 0.064.
$$
However, there is some uncertainty in this estimate of $\lambda$, but more fundamentally, the conditional probability distribution of the number of accidents in the eight-day period, given that the number of accidents in $122$ days was $500$ does not actually depend on $\lambda$.  It is the binomial distribution of the number of successes in $500$ trials with probability $122/500$ of success on each trial.  This is a problem involving a binomial distribution.
$$
\binom{500}{30} \left( \frac 8 {122} \right)^{30} \left( \frac{122-8}{122} \right)^{500-30} \approx 0.065767.
$$
Theorem. Suppose $X,Y$ are independent and $X\sim\mathrm{Poisson}(\alpha)$ and $Y\sim\mathrm{Poisson}(\beta)$.  Then the conditional distribution of $X$ given $X+Y$ is $\mathrm{Binomial} \left( X+Y, \dfrac\alpha{\alpha+\beta} \right)$.
In this case $X$ is the number of accidents in the eight-day period and $Y$ is the number of accidents in the remaining $122-8$ days.
