How Do Poisson and Binomial Distributions Apply to Wire Flaw Analysis?

[skrat]

Homework Statement

We assume that the number of structural flaws on a long wire have obey Poisson distribution law. On average we find 1 flaw every 5 meters.

a) What is the probability that a 20 m long section will have maximum 2 flaws?
b) We slice the wire into 1 m long sections. What is the probability that 3 or less sections (out of 10) have one flaw or more?

Homework Equations

The Attempt at a Solution

a) $$P=e^{-0.2\cdot 20}(1+0.2\cdot 20+\frac{(0.2\cdot 20)^2}{2})=0.238$$
b) Probability that there is NO mistake on a 1m long section is $$P=1-e^{-0.2}=0.18$$ now using Binomial distribution the probability should be $$P=1-\sum _{i=0}\binom{10}{i}(1-0.18)^i0.18^{10-i}$$ yet the results I have say that $$P=1-\sum _{i=0}\binom{10}{i}0.18^{i}(1-0.18)^{10-i}.$$

I personally disagree with that "official" result but would like to hear your opinion...!

 

[skrat]

Ah, forget it.

b) The probability 0.18 applies to that there is at least one (or more) flaw on a 1 m section.