Probability of making a claim

For an individual, the probability of having an accident in a period of 24 hrs (and therefore the probability of making a claim) is 0.00037. Claims on successive days are independent, and a person can not have more than two accidents in one day. Calculate the probability that an insured does 0, 1 and 2 claims in a year.

I will suppose that every time that somebody has an accident, that person makes a claim. I was pretty sure this is Binomial distribution, because each day can be seen as a trial, and having an accident a success (success on making a claim). But then I read again the problem, and it says that "a person can not have more than two accidents in one day", so a day can be seen as one of the trials, so maybe I was thinking of instead of making n= 365, making it equal to 730 (which is 365 times 2). Because one can have 2 accidents a day maximum.

Can somebody help me out?

• I suspect that they meant "cannot have $2$ or more accidents in one day", because they only gave the probability of having an accident in a 24 hour period. – Graham Kemp Nov 12 '14 at 2:14
• But, yes, the count of accidents per year has a binomial probability distribution. – Graham Kemp Nov 12 '14 at 2:17
• So there's no possible way to solve this? It was really bothering me. Thanks! – Di Alejandra Nov 12 '14 at 3:18

Use $\mathsf P(X=x) = {365\choose x}0.00037^x \,0.99963^{365-x}$ and substitute for $x\in\{0,1,2\}$