Poisson distribution help? The number of car accidents occurring per day on a highway follows a Poisson distribution with mean 1.5.
a) Given that at least one accident occurs on another day, find the probability that more than two accidents occur on that day.
b) What is the minimum number of days that would pass in order to have a probability of 95% that at least one accident occurs?
Edit:
For a) I've worked that the probability of at least one is .442175, and the probability of two or more is .77867, but I'm not sure what to do with those two numbers? Should I divide one of them by the other?
For b) our teacher hinted that this question is meant to move towards a new distribution technique? I'm not sure how many cars crash each day. 
 A: Now that you have indicated your work, I can write out a solution to the problem. Recall that if $A$ and $B$ are events, with $\Pr(B)\ne 0$, then the conditional probability $\Pr(A|B)$ of $A$ given $B$ is given by
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
Let $B$ be the event "at least one accident" and let $A$ be the event "more than two accidents."
We have $\Pr(B)=1-e^{-\lambda}$, where $\lambda=1.5$.
The event $A\cap B$ occurs if there are more than $2$ accidents. Its probability is $1$ minus the probability of $2$ or fewer accidents. It follows that 
$$\Pr(A\cap B)=1-e^{-\lambda}\left(1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}\right).$$
Finally, divide, as in Equation (1).
For b), use the fact that the number of accidents in time $t$ has Poisson distribution with parameter $\lambda t$, where $\lambda$ is the mean number of accidents per unit time, here $1.5$. The probability of at least one accident is $1-e^{-\lambda t}$. Set this equal to $0.95$.
We get $e^{-\lambda t}=0.05$. Solve for $t$, by taking the natural logarithm of both sides.
Remark: I feel a little guilty about asserting without proof the fact about the distribution of the number of accidents in $t$ days. For our problem, we do not need the full fact. Suppose that the numbers of accidents from day to day are independent random variables. Then, since the probability of no accidents in $1$ day is $e^{-\lambda}$,  the probability of no accidents in $n$ days, where $n$ is a positive integer is $(e^{-\lambda})^n$, which is $e^{-\lambda n}$. 
It is reasonable to believe (and true) that even if $t$ is not an integer, the probability of no accidents in $t$ days is $(e^{-\lambda})^t$. 
