Question

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Solution needed with a fair explanation

Answer 1

Answer: c. 0.40396

Explanation:

The total number of flaws is given by:

`{\left({0}*{4}\right)}+{\left({1}*{3}\right)}(0*4)+(1*3) +{\left({2}*{5}\right)}+{\left({3}*{2}\right)}+(2*5)+(3*2) +{\left({4}*{4}\right)}+{\left({5}*{1}\right)}+(4*4)+(5*1) +{\left({6}*{1}\right)}+(6*1)={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<3)`

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

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

Hence, c. is the correct option.