virus is spreading in a city. 6% of households who have pets will be affected, while 1% of households without pets will be affected. In this city, about 20% of the households have pets and 80% don't. Conditioning on a household is affected by this virus, what is the probability that this household has a pet?
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2What you're looking for is Bayes' theorem. Have a look, try to apply it, and if it still doesn't work, then you can tell us where you got stuck and we will try to help you. – Arthur Nov 21 '18 at 15:43
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I know that this is Bayes' formula application but I got stuck to apply it to this condition – Mohamed Medhat Ali Nov 21 '18 at 16:11
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Where did you get stuck? How did you try to apply it? Please edit your post and tell us more so we can help you where you are. – Arthur Nov 21 '18 at 16:17
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% of affected households with pets divided by the percentage of households that are affected . Do I need to multiply 0.2 * 0.06 and 0.01 * 0.8 ? and by Bayes' formula I try this (0.060.20)/[(0.060.20)+(0.01*0.80) – Mohamed Medhat Ali Nov 21 '18 at 16:37
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We should formulate the problem to be able to attain the answer: $$\Pr(\text{Being affected}|\text{Having pets})=6\%\\\Pr(\text{Being affected}|\text{Not having pets})=1\%\\\Pr(\text{Having pets})=20\%\\$$Then we are looking for$$\Pr(\text{Having pets}|\text{Being affected})$$which using Bayes's theorem can be calculated as$$\Pr(\text{Having pets}|\text{Being affected})\\=\\{\Pr(\text{Having pets})\Pr(\text{Being affected}|\text{Having pets})\over\Pr(\text{Having pets})\Pr(\text{Being affected}|\text{Having pets})+\Pr(\text{Not having pets})\Pr(\text{Being affected}|\text{Not having pets})}\\=\\{20\%\times 6\%\over 20\%\times 6\%+80\%\times 1\%}=60\%$$
Mostafa Ayaz
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