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Question Number 96730 by Ar Brandon last updated on 04/Jun/20

$$\mathcal{A}\mathrm{third}\mathrm{of}\mathrm{a}\mathrm{population}\mathrm{has}\mathrm{been}\mathrm{vaccined}\mathrm{against}$$$$\mathrm{an}\mathrm{illness}.\mathcal{D}\mathrm{uring}\mathrm{the}\mathrm{pandemie}\mathrm{it}\mathrm{is}\mathrm{noticed}\mathrm{that};$$$$\mathrm{out}\mathrm{of}15\mathrm{patients},2\mathrm{have}\mathrm{been}\mathrm{vaccined}.$$$$\mathcal{A}\mathrm{ssuming}\mathrm{that}\mathrm{out}\mathrm{of}\mathrm{a}\mathrm{hundred}\mathrm{vaccined},8\mathrm{are}\mathrm{ill}.$$$$\mathcal{A}\mathrm{n}\mathrm{individual}\mathrm{is}\mathrm{choosen}\mathrm{at}\mathrm{random}\mathrm{from}\mathrm{this}\mathrm{population}.$$$$\mathcal{L}\mathrm{et}\mathbf{M}\mathrm{imply}\mathbf{the}\mathbf{individual}\mathbf{is}\mathbf{ill}\mathrm{and}\mathbf{V}\mathrm{imply}\mathbf{the}\mathbf{individual}$$$$\mathbf{has}\mathbf{been}\mathbf{vaccined}.$$$$1\mathrm{\setminus }\mathcal{D}\mathrm{etermine}\mathrm{the}\mathrm{probabilities}:\mathrm{P}\left(\mathrm{V}\right);\mathrm{P}\left(\mathrm{V}/\mathrm{M}\right)\mathrm{and}\mathrm{P}\left(\mathrm{M}/\mathrm{V}\right)$$$$2\mathrm{\setminus }\mathcal{C}\mathrm{alculate}\mathrm{P}(\mathrm{M}\cap \mathrm{V})\mathrm{then}\mathrm{P}\left(\mathrm{M}\right).\mathcal{D}\mathrm{educe}\mathrm{the}$$$$\mathrm{percentage}\mathrm{of}\mathrm{patients}.$$

Answered by Rio Michael last updated on 04/Jun/20

$$\left(1\right)P\left(V\right)=P(V\cap M^{\prime })\cup P(V\cap M)=\left(\frac{1}{3}\right)\left(\frac{2}{15}\right)\left(\frac{92}{100}\right)+\left(\frac{1}{3}\right)\left(\frac{13}{15}\right)\left(\frac{92}{100}\right)$$$$P\left(V/M\right)=\frac{P(V\cap M)}{P\left(M\right)}=\frac{\left(\frac{1}{3}\right)\left(\frac{13}{15}\right)\left(\frac{92}{100}\right)}{\left(\frac{1}{3}\right)\left(\frac{2}{15}\right)\left(\frac{8}{100}\right)+\left(\frac{2}{3}\right)\left(\frac{13}{15}\right)\left(\frac{8}{100}\right)}$$$$P\left(M/V\right)=\frac{P(M\cap V)}{P\left(V\right)}=\frac{\left(\frac{1}{3}\right)\left(\frac{13}{15}\right)\left(\frac{92}{100}\right)}{\left(\frac{1}{3}\right)\left(\frac{2}{15}\right)\left(\frac{92}{100}\right)+\left(\frac{1}{3}\right)\left(\frac{13}{15}\right)\left(\frac{92}{100}\right)}$$$$\left(2\right)P(M\cap V)=\left(\frac{1}{3}\right)\left(\frac{13}{15}\right)\left(\frac{92}{100}\right)\mathrm{then}P\left(M\right)=\mathrm{see}P\left(V/M\right)$$$$\mathrm{\%}P\left(M\right)=\frac{P\left(M\right)}{P\left(V\right)}\times 100$$

Commented by Rio Michael last updated on 04/Jun/20

$$\mathrm{please}\mathrm{check}\mathrm{this}\mathrm{solution},\mathrm{i}\mathrm{may}\mathrm{have}\mathrm{made}\mathrm{mistake}$$

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