Question

If \(3 \%\) of electronic units manufactured by a company are defective. Find the probability that in a sample of 200 units, less than 2 bulbs are defective.

Answer 1

The probability of defective units \(p=3 / 100=0.03\)
Give \(n=200\).
We observe that \(p\) is small and \(n\) is large here. Thus it is a Poisson distribution.
Mean \(\lambda=n p=200 \times 0.03=6\)
\(P(X=x)\) is given by the Poisson Distribution Formula as \(\left(e^{-\lambda} \lambda^{x}\right.\) ) \(/ x !\)
\(P(X<2)=P(X=0)+P(X=1)\)
\(=\left(e^{-6} 6^{0}\right) / 0 !+\left(e^{--6} 6^{1}\right) / 1 !\)
\(=e^{-6}+e^{-6} \times 6\)
\(=0.00247+0.0148\)
\(P(X<2)=0.01727\)