Question

the number of defective production in a production process follows a poisson distribution with a mean of 2.6 per month, for a given month what is the probability there will be fewer than two defective production?

Answer 1

Let the random variable $X$ represent the number of defective productions in a production process the $X\sim Poisson(\lambda=2.6)$ given as,

$p(X=x)={e^{-2.6}2.6^x\over x!}, \space x=0,1,2,3,.. \\0,\space elsewhere$

We find the probability,

$p(X\lt2)=p(X=0)+p(X=1)$

Now,

$p(X=0)={e^{-2.6}2.6^0\over 0!}=0.07427358$ and $p(X=1)={e^{-2.6}2.6^1\over 1!}=0.19311131$. Therefore, $p(X\lt 2)=0.07427358+0.19311131=0.26738489$

Thus, the probability that for a given month there will be fewer than two defective productions is 0.26738489