# 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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\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}