# Thread: Probability & Poisson Theory

1. ## Probability & Poisson Theory

It is my understanding that the Poisson Theory applies in the following question but I still require help -

Given a fixed period of time what is the probablity of any number of sales for each product given that the expected total sales on average is 3.

Product A most sales 60%
Product B most sales 15%
Equal Sales 25%

Expected sales
3

Knowing these values, what is the probability of

Product A 0 sales & Product B 0 sales

Product A 1 sale & Product B 1 sale

Product A 1 sale & Product B 0 sales

Product A 3 sales & Product B 1 sale

and so on etc

When all variations of sales (with the most that could possibly be sold is 8) are calculated they should equal the corresponding expected sales i.e 0-0, 1-1, 2-2. 3-3, 4-4 should be the same as Equal Sales of 25% and the same for the others.

I hope i have explained the problem well enough and would appreciate any helpful comments or answers.

I have been using excel to calcuate this but there appear to be a number of inconsistancies when any of the starting values are changed.

2. Your paraphrasing makes it a lot harder to understand the problem.

During a fixed period of time, the expected total sales for two different products is 3 items. The probability that product A has the most sales is 30%, that product B has the most sales is 15%, and that there are as many sales of product A as B is 25%.

No more than 8 could be sold? How is that? I don't see how you could use Poisson if you have a cap of 8. Poisson distributions have infinitely many outcomes. Are there just 4 of each product? You could try to apply Poisson, but there will be non-zero probability of selling more than 8 items.

Are you sure you didn't mean some other distribution?