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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.
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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 $\left(e^{-\lambda} \lambda^{x}\right.$ ) $/ x !$
$P(X<2)=P(X=0)+P(X=1)$
$=\left(e^{-6} 6^{0}\right) / 0 !+\left(e^{--6} 6^{1}\right) / 1 !$
$=e^{-6}+e^{-6} \times 6$
$=0.00247+0.0148$
$P(X<2)=0.01727$
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