Thread: Poisson distribution

Well I don't really know if the question I am going to put in here is a pre-University question or if I'm posting in the right place but anyway here it goes:
Consider the following exercice:
"The number of flaws in bolts of cloth in textile manufacturing
is assumed to be Poisson distributed with a mean of
0.1 flaw per square meter.
What is the probability that there is one flaw in 10 square
meters of cloth?"

Well the book says the way to solve it is to say, well if mean value is 0.1 per square meter then it is 1 per 10 square meters and then it's a matter of substituting the values in the poison distribution expression. However, I thought it could be solved other way, I thought:having one in 10 square meters is the same as having one in the first meter and none in the other nine! or one in the second and none in the others, etc... It means then that the probability of having one in 10 meters is ten times the probabiblity of having one in the first times (the probability of having none in one meter)^9 but that gives me a different answer... what is wrong here? It can't be an issue about the independence of each partition cause so a random variable can obey a poison distributions it's a must to have independence between partitions isn't it? can anyone tell me? thanks!

why don't you show your work on how there is none in the other squares.
The error most likely is in that calculation.