Question

Defects in a certain wire occur at the rate of one per 10 meter. Assume the defects have a Poisson distribution. What is the probability that: (a) a 20-meter wire has no defects? (b) a 20-meter wire has at most 2 defects?

Answer 1

We have a Poisson distribution,

$λ=1 \text{ defects per 10 meter};\\ \text{for 20-meter wire } t = 2;\\ P_t(X=k)=\cfrac{(\lambda t)^k\cdot e^{-\lambda t}}{k!}=\cfrac{(1\cdot2)^k\cdot e^{-1\cdot2}}{k!}=\\ =\cfrac{2^k\cdot e^{-2}}{k!};\\ \text{(a) }P_2(X=0)=\cfrac{2^0\cdot e^{-2}}{0!}=0.1353.\\ \text{(b) } P_2(X\le2)=\\ =P_2(X=0)+P_2(X=1)+P_2(X=2)=\\ =\cfrac{2^0\cdot e^{-2}}{0!}+\cfrac{2^1\cdot e^{-2}}{1!}+\cfrac{2^2\cdot e^{-2}}{2!}=\\ =0.1353+0.2707+0.2707=0.6767.$