91Ó°ÊÓ

A magnetic tap contains, on average 3 defects per 1000 feet. We have to determine the probability that a roll of the tap 1200 feet long contains no defects.

Let X be the Poisson variable that denotes a roll of tape 1200 feet long and contains no defects.

The magnetic tape contains, on average 3 defects per 1000 feet. So, the probability of a foot will be defective is \(\frac{3}{{1000}} = 0.003\).So in a roll of 1200 feet long, the expected number of defectives is 0.003\( \times \)1200=3.6 \(\left( \lambda \right)\).

So, X follows a Poisson distribution with parameter 3.6

We know that

\(\begin{array}{l}P\left( {X = x} \right) = \frac{{{\lambda ^x}}}{{x!}}{e^{ - \lambda }}\\P\left( {X = 0} \right) = \frac{{{{\left( {3.6} \right)}^0}}}{{0!}}{e^{ - 3.6}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {e^{ - 3.6}}\end{array}\)