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could you help me solve this : Conditioning on a household is affected by this virus, what is the probability...
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?
conditional-probability
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
add a comment |
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?
conditional-probability
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
2
What 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
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
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
% 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 (0.06*0.20)/[(0.06*0.20)+(0.01*0.80)
– Mohamed Medhat Ali
Nov 21 '18 at 16:37
add a comment |
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?
conditional-probability
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
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?
conditional-probability
conditional-probability
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
asked Nov 21 '18 at 15:40
Mohamed Medhat Ali
31
31
2
What 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
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
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
% 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 (0.06*0.20)/[(0.06*0.20)+(0.01*0.80)
– Mohamed Medhat Ali
Nov 21 '18 at 16:37
add a comment |
2
What 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
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
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
% 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 (0.06*0.20)/[(0.06*0.20)+(0.01*0.80)
– Mohamed Medhat Ali
Nov 21 '18 at 16:37
2
2
What 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
What 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
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
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
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
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
% 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 (0.06*0.20)/[(0.06*0.20)+(0.01*0.80)
– Mohamed Medhat Ali
Nov 21 '18 at 16:37
% 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 (0.06*0.20)/[(0.06*0.20)+(0.01*0.80)
– Mohamed Medhat Ali
Nov 21 '18 at 16:37
add a comment |
1 Answer
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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})overPr(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%$$
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
We should formulate the problem to be able to attain the answer:
$$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$$Pr(text{Having pets}|text{Being affected})\=\{Pr(text{Having pets})Pr(text{Being affected}|text{Having pets})overPr(text{Having pets})Pr(text{Being affected}|text{Having pets})+Pr(text{Not having pets})Pr(text{Being affected}|text{Not having pets})}\=\{20%times 6%over 20%times 6%+80%times 1%}=60%$$
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
add a comment |
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})overPr(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%$$
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
add a comment |
We should formulate the problem to be able to attain the answer:
$$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$$Pr(text{Having pets}|text{Being affected})\=\{Pr(text{Having pets})Pr(text{Being affected}|text{Having pets})overPr(text{Having pets})Pr(text{Being affected}|text{Having pets})+Pr(text{Not having pets})Pr(text{Being affected}|text{Not having pets})}\=\{20%times 6%over 20%times 6%+80%times 1%}=60%$$
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
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})overPr(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%$$
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
answered Nov 21 '18 at 17:09
Mostafa Ayaz
14.1k3937
14.1k3937
add a comment |
add a comment |
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2
What 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
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
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
% 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 (0.06*0.20)/[(0.06*0.20)+(0.01*0.80)
– Mohamed Medhat Ali
Nov 21 '18 at 16:37