# Find next month

I have a man that sells in average 36 deal papers every year.I am searching the possibility to sell more than 2 deal papers,in the next mouth. I take this and solve it $$f(x)={n \choose x} p^x (1-p)^{n-x}$$ For n=36 x=2 p=30 (from months) I solve it and i find 4 is this right?A friend of me told me it isn't Dionymic nor Poisson

• You have to know which distribution this is. Otherwise, you can only get bounds, no exact value. – астон вілла олоф мэллбэрг Sep 11 at 12:02
• how can i know?it confuses me the next mouth .. – m.s Sep 11 at 12:03
• I would assume that the correct distribution here to use is poisson. It certainly isn't likely to be binomial like your attempt is. And, "I solve it and I find 4" Surely you aren't thinking that $4$ is the correct answer to a probability question... – JMoravitz Sep 11 at 12:18
• wow really?could you explain me why it is Poisson? – m.s Sep 11 at 12:35
• Because it seems plausible that the man might have sold a number of papers different than $36$ in the year. If he gets particularly lucky, maybe he sold 50 papers. Maybe if he got particularly unlucky he only sold five papers total in a year. It seems plausible that if the right bit of luck occurs that he sells any number of papers in the year from $0,1,2,3,\dots$ on up into the trillions. Granted... in reality he won't be selling billions of papers, there wouldn't be enough resources or people to sell to... but the point is that he isn't guaranteed to sell exactly $36$ next year. – JMoravitz Sep 11 at 13:08