Question

3.48. The nulliber of flaws in a disk follows a Poisson distribution with the mean 0.5, Π 5 disks are selected independently, what is the probability that none of the disks have flaws? What is the probability that all 5 have a total of 2 flaws? The total of at least 2 flaws?

Answer 1

$Let~X~be~random~variable~denoting~no.~of~flaws~in~a~disk.\\ X\sim Poisson(0.5).\\ PMF~of~X~is\\ P(X=x)=Probability~that~no.~of~flaws~in~a~disk~is~x\\ =\frac{e^{-0.5}(0.5)^x}{x!};~~x=0,1,2,....\\ =0~otherwise\\ Probability~that~none~of~the~disks~have~flaws=(P(X=0))^5\\ =e^{-2.5}=0.0821\\ Since~the~flaw~in~a~disk~does~not~depend~on~other~disks~hence~independent.\\ Probability~that~all~5~have~a~total~of~2~flaws\\ =Probability~that~any~one~of~5~exactly~two~flaws~and~remaining~have~no~flaw\\ +Probability~that~any~two~of~five~exactly~one~flaw~and~remaining~have~no\\ flaws=\binom{5}{1}((0.5)^2*e^{-0.5}/2!)*(e^{-0.5})^4+\binom{5}{2}((0.5)*e^{-0.5})^2*(e^{-0.5})^3\\ =(5*0.25/2+10*0.25)*exp(-2.5)=0.2565\\ Probability~that~the~total~of~at least~2~flaws\\=1-Probability~that~total~of~at~most~one~flaws\\ =1-Probability~that~none~of~the~disks~have~flaws\\+Probability~that~total~of~exactly~one~flaw\\ =1-(e^{-0.5})^5-(5*0.25/2 )(e^{-0.5})^5=0.8666$