So I'm just unsure how I set up my answer to the following question, which is a homework question. " The number of surfaces blemished on any one chocolate is described by a Poisson distribution with parameter= 0.05. What is the probability that a box of 10 chocolates will contain two or more chocolate each with one or more surfaces blemishes?"
So P= Probability. My initial thoughts were that this going to be simple 1-(P(0 blemished)+P(1 blemished). But whats really troubling me is the last condition "each with one or more surfaces blemishes". Does this mean there is another distribution I must account for which has amount of surfaces blemished? Or maybe I can do it as binomial distribution with 1- [10 C 1 e^(-0.05) *( 1- e^(-0.05))^9] . Would this work? Don't expect anyone to do whole question but can you please tell me if I'm thinking the right way about it? Really getting confused about how i break this down and structure a solution
The problem has two steps. In the first place you need to know the probability that any given single chocolate is blemished. That is given by $1-F(0)$ where $F(n)$ is the Poisson CDF with $\mu = .05$. Call this answer $p$; it will be a bit smaller than $.05$.
The second step is that now you have ten Bernoulli variates with $p$ equal to the $p$ described above, and you need to find the proability that 2 or more will be "true". This is easiest to do by subtracting (from 1) the probablity that none will hit (which is $(1-p)^{10}$), and the probablity that one will hit (which is $p^1(1-p)^{9}\binom{10}{1}$).
So you were on the right track with your binomial calculation.