# 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!

• "the total number of accidents in the eight months" Might you have meant "eight days" rather than "eight months"? Commented Dec 25, 2015 at 19:22

I would say there are $500$ accidents in $31+30+31+30=122$ days, so $\lambda=500/122$.

• this is a λ for 1 day, right ? so I need to multiply it by 8. Commented Dec 25, 2015 at 19:27
• Yes, for one day. Commented Dec 25, 2015 at 19:29
• isn't it the same thing that I did, except that you took two days with 31 days? Commented Dec 25, 2015 at 19:29
• I was worried how you got 100/3, or 33 crashes a day? Mine is closer to 4 crashes a day. Commented Dec 25, 2015 at 19:31
• I averaged the number of accidents a month, to get 125 a month, and 100/3 is for 8 days. Is it correct then that the probability is 0.064289 ? Commented Dec 25, 2015 at 19:33

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.