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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.
by Diamond (89,175 points) | 1,648 views

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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 \(\left(e^{-\lambda} \lambda^{x}\right.\) ) \(/ x !\)
\(=\left(e^{-6} 6^{0}\right) / 0 !+\left(e^{--6} 6^{1}\right) / 1 !\)
\(=e^{-6}+e^{-6} \times 6\)
by Diamond (89,175 points)

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