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 \(\left(e^{-\lambda} \lambda^{x}\right.\) ) \(/ x !\)

\(P(X<2)=P(X=0)+P(X=1)\)

\(=\left(e^{-6} 6^{0}\right) / 0 !+\left(e^{--6} 6^{1}\right) / 1 !\)

\(=e^{-6}+e^{-6} \times 6\)

\(=0.00247+0.0148\)

\(P(X<2)=0.01727\)