Question

Spammers will be reported! Solution needed with a fair explanation

Answer 1

Answer: c. 0.40396

Step-by-step explanation:

The total number of flaws is given by:

${\left({0}\times{4}\right)}+{\left({1}\times{3}\right)}(0\times4)+(1\times3) +{\left({2}\times{5}\right)}+{\left({3}\times{2}\right)}+(2\times5)+(3\times2) +{\left({4}\times{4}\right)}+{\left({5}\times{1}\right)}+(4\times4)+(5\times1) +{\left({6}\times{1}\right)}+(6\times1)={46}=46$

Average number of flaws for the 20 sheets : $\displaystyle\mu=\frac{46}{{20}}={2.3}$

Required probability : $=P(X\geq3)=1-P(X$

$=1-[P(X=0)+P(X=1)+P(X=2)]$

$=1-(\dfrac{e^{-2.3}2.3^0}{0!}+\dfrac{e^{-2.3}2.3^1}{1!}+\dfrac{e^{-2.3}2.3^2}{2!})\ \ \ [\text{Poisson distribution formula: }P(X=x)=\dfrac{e^{-\mu}\mu^x}{x!}]\\\\=1-0.596038825932=0.403961174068\approx0.40396$

Hence, c. is the correct option.