Question

A machine producing hair pins produces 1 defective out of 400 on an average. If 100 hairpins are packed in each box.

In above problem, suppose the manufacturer gives guarantee that there will be less than 3 defectives in a box of 100. What is the probability that a box selected at random will meet the guarantee.

Answer 1

$n=100, p=1/400=0.0025.$

We have a Bernoulli trial - exactly two possible outcomes, "success" (the hairpin is defective) and "failure" (the hairpin is non-defective) and the probability of success is the same every time the experiment is conducted (a hairpin is produced).

The probability of each result

$P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}\cdot p^k\cdot q^{n-k}.$

For those situations in which $n$ is large and $p$ is very small, the Poisson distribution can be used to approximate the binomial distribution.

We have a Poisson distribution,

$\lambda=np=100\cdot0.0025=0.25;\\ P(X=k)=\cfrac{\lambda^k\cdot e^{-\lambda}}{k!}.$

$P(X<3) =\\ =P(X=0)+P(X=1)+P(X=2)=\\ =\cfrac{0.25^0\cdot e^{-0.25}}{0!}+\cfrac{0.25^1\cdot e^{-0.25}}{1!}+\cfrac{0.25^2\cdot e^{-0.25}}{2!}=\\ =0.7788+0.1947+0.0243=0.9978.$