# Using a Poisson distribution, how do find if 'X' or more of 'Y' trials pass?

I am presented with the questions:

Flaws occur in mylar material according to a Poisson distribution with a mean of 0.05 flaw per square yard.

(a) If 29 square yards are inspected, what is the probability that there are no flaws? (b) What is the probability that a randomly selected square yard has no flaws? (c) Suppose that the mylar material is cut into 5 pieces, each being 1 yard square. What is the probability that 3 or more of the 5 pieces will have no flaws?

I know:

a) poissonpdf(29*.05,0) = .235

b) poissonpdf(.05,0) = .951

My problem is with part 'c'. I'm completely stumped on how to get the answer. I know it uses poissoncdf, but I can't get my numbers to make sense. Can anyone explain to me how I might go about getting the answer?

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The probability that a square yard of material has no flaws is, from your b), 0.951. Or alternatively, the probability that a given yard has one or more flaws is 0.049.

The probability that 3 or more of 5 have no flaws is a Binomial distribution. Alternatively, that 0, 1 or 2 have a flaw. From the cdf with a probability of 0.049, this is 0.999.

To explain more as requested:

\begin{align} P(X<=2)&=P(X=0)+P(X=1)+P(X=2)\\ &= {5\choose0}0.049^00.951^5+{5\choose1}0.049^10.951^4+{5\choose2}0.049^20.951^3 \end{align}

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That is correct, but can you explain a bit more how you got that .999 answer? –  rphello101 Sep 19 '13 at 3:48