Trouble understanding Bayes Theorem

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I was watching a video on YouTube and i am not sure if the given solution is correct. Can someone confirm?



answe to problem unsure if correct






probability naive-bayes







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  • The final + in the denominator should be x.
    – BruceET
    6 hours ago






  • 5




    There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
    – Tim♦
    6 hours ago
















up vote
1
down vote

favorite
1












I was watching a video on YouTube and i am not sure if the given solution is correct. Can someone confirm?



answe to problem unsure if correct






probability naive-bayes







share|cite|improve this question























  • The final + in the denominator should be x.
    – BruceET
    6 hours ago






  • 5




    There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
    – Tim♦
    6 hours ago












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I was watching a video on YouTube and i am not sure if the given solution is correct. Can someone confirm?



answe to problem unsure if correct






probability naive-bayes







share|cite|improve this question















I was watching a video on YouTube and i am not sure if the given solution is correct. Can someone confirm?



answe to problem unsure if correct






probability naive-bayes




probability naive-bayes






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edited 21 mins ago

























asked 7 hours ago









siddhartha pachhai

435




435











  • The final + in the denominator should be x.
    – BruceET
    6 hours ago






  • 5




    There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
    – Tim♦
    6 hours ago
















  • The final + in the denominator should be x.
    – BruceET
    6 hours ago






  • 5




    There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
    – Tim♦
    6 hours ago















The final + in the denominator should be x.
– BruceET
6 hours ago




The final + in the denominator should be x.
– BruceET
6 hours ago




5




5




There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
6 hours ago




There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
6 hours ago










2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Seems correct, except for the typo noted in my Comment. Let $D$ indicate 'has disease' and $T$ indicate 'tests positive'.



Bayes' Theorem states the following (denoting intersection of events as 'multiplication'):



$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD)+P(D^c)P(T.$$



Sometimes, $P(T) = P(D)P(T|D)+P(D^c)P(T|D^c)$ is called the Law of Total Probability.



You are given that $P(D) = 0.01,, P(T|D) = P(T^c|D^c) = 0.99.$
Then by the Complement Rule, $P(D^c) = 0.99,, P(T|D^c) = 0.01.$
Plug in these numbers to get the answer claimed.



To understand these probabilities, notice that $P(DT), P(D|T),$ and $P(T|D)$ all contain the same events, but they refer to three different populations:
the first to the population of everyone who may take the test, the second to
the population of those who tested positive, and the third to the population of those who have the STD.



Note: You can find more detailed discussions of this kind of situation in several
of the links under 'Related' in the right margin. Also, you may want to look
at Wikipedia articles on "Bayes' Theorem" and "screening test."






share|cite|improve this answer






















  • Okay, yea thank you!
    – siddhartha pachhai
    11 mins ago


















up vote
2
down vote













Yes even without consulting the equations it is possible to work it out from the information. See below.
Work it out






share|cite|improve this answer








New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Thanks, this method also explains alot
    – siddhartha pachhai
    8 mins ago










Your Answer




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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Seems correct, except for the typo noted in my Comment. Let $D$ indicate 'has disease' and $T$ indicate 'tests positive'.



Bayes' Theorem states the following (denoting intersection of events as 'multiplication'):



$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD)+P(D^c)P(T.$$



Sometimes, $P(T) = P(D)P(T|D)+P(D^c)P(T|D^c)$ is called the Law of Total Probability.



You are given that $P(D) = 0.01,, P(T|D) = P(T^c|D^c) = 0.99.$
Then by the Complement Rule, $P(D^c) = 0.99,, P(T|D^c) = 0.01.$
Plug in these numbers to get the answer claimed.



To understand these probabilities, notice that $P(DT), P(D|T),$ and $P(T|D)$ all contain the same events, but they refer to three different populations:
the first to the population of everyone who may take the test, the second to
the population of those who tested positive, and the third to the population of those who have the STD.



Note: You can find more detailed discussions of this kind of situation in several
of the links under 'Related' in the right margin. Also, you may want to look
at Wikipedia articles on "Bayes' Theorem" and "screening test."






share|cite|improve this answer






















  • Okay, yea thank you!
    – siddhartha pachhai
    11 mins ago















up vote
2
down vote



accepted










Seems correct, except for the typo noted in my Comment. Let $D$ indicate 'has disease' and $T$ indicate 'tests positive'.



Bayes' Theorem states the following (denoting intersection of events as 'multiplication'):



$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD)+P(D^c)P(T.$$



Sometimes, $P(T) = P(D)P(T|D)+P(D^c)P(T|D^c)$ is called the Law of Total Probability.



You are given that $P(D) = 0.01,, P(T|D) = P(T^c|D^c) = 0.99.$
Then by the Complement Rule, $P(D^c) = 0.99,, P(T|D^c) = 0.01.$
Plug in these numbers to get the answer claimed.



To understand these probabilities, notice that $P(DT), P(D|T),$ and $P(T|D)$ all contain the same events, but they refer to three different populations:
the first to the population of everyone who may take the test, the second to
the population of those who tested positive, and the third to the population of those who have the STD.



Note: You can find more detailed discussions of this kind of situation in several
of the links under 'Related' in the right margin. Also, you may want to look
at Wikipedia articles on "Bayes' Theorem" and "screening test."






share|cite|improve this answer






















  • Okay, yea thank you!
    – siddhartha pachhai
    11 mins ago













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Seems correct, except for the typo noted in my Comment. Let $D$ indicate 'has disease' and $T$ indicate 'tests positive'.



Bayes' Theorem states the following (denoting intersection of events as 'multiplication'):



$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD)+P(D^c)P(T.$$



Sometimes, $P(T) = P(D)P(T|D)+P(D^c)P(T|D^c)$ is called the Law of Total Probability.



You are given that $P(D) = 0.01,, P(T|D) = P(T^c|D^c) = 0.99.$
Then by the Complement Rule, $P(D^c) = 0.99,, P(T|D^c) = 0.01.$
Plug in these numbers to get the answer claimed.



To understand these probabilities, notice that $P(DT), P(D|T),$ and $P(T|D)$ all contain the same events, but they refer to three different populations:
the first to the population of everyone who may take the test, the second to
the population of those who tested positive, and the third to the population of those who have the STD.



Note: You can find more detailed discussions of this kind of situation in several
of the links under 'Related' in the right margin. Also, you may want to look
at Wikipedia articles on "Bayes' Theorem" and "screening test."






share|cite|improve this answer














Seems correct, except for the typo noted in my Comment. Let $D$ indicate 'has disease' and $T$ indicate 'tests positive'.



Bayes' Theorem states the following (denoting intersection of events as 'multiplication'):



$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD)+P(D^c)P(T.$$



Sometimes, $P(T) = P(D)P(T|D)+P(D^c)P(T|D^c)$ is called the Law of Total Probability.



You are given that $P(D) = 0.01,, P(T|D) = P(T^c|D^c) = 0.99.$
Then by the Complement Rule, $P(D^c) = 0.99,, P(T|D^c) = 0.01.$
Plug in these numbers to get the answer claimed.



To understand these probabilities, notice that $P(DT), P(D|T),$ and $P(T|D)$ all contain the same events, but they refer to three different populations:
the first to the population of everyone who may take the test, the second to
the population of those who tested positive, and the third to the population of those who have the STD.



Note: You can find more detailed discussions of this kind of situation in several
of the links under 'Related' in the right margin. Also, you may want to look
at Wikipedia articles on "Bayes' Theorem" and "screening test."







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 6 hours ago

























answered 6 hours ago









BruceET

3,7281519




3,7281519











  • Okay, yea thank you!
    – siddhartha pachhai
    11 mins ago

















  • Okay, yea thank you!
    – siddhartha pachhai
    11 mins ago
















Okay, yea thank you!
– siddhartha pachhai
11 mins ago





Okay, yea thank you!
– siddhartha pachhai
11 mins ago













up vote
2
down vote













Yes even without consulting the equations it is possible to work it out from the information. See below.
Work it out






share|cite|improve this answer








New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Thanks, this method also explains alot
    – siddhartha pachhai
    8 mins ago














up vote
2
down vote













Yes even without consulting the equations it is possible to work it out from the information. See below.
Work it out






share|cite|improve this answer








New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Thanks, this method also explains alot
    – siddhartha pachhai
    8 mins ago












up vote
2
down vote










up vote
2
down vote









Yes even without consulting the equations it is possible to work it out from the information. See below.
Work it out






share|cite|improve this answer








New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









Yes even without consulting the equations it is possible to work it out from the information. See below.
Work it out







share|cite|improve this answer








New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 6 hours ago









Curtis White

1212




1212




New contributor




Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Curtis White is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Thanks, this method also explains alot
    – siddhartha pachhai
    8 mins ago
















  • Thanks, this method also explains alot
    – siddhartha pachhai
    8 mins ago















Thanks, this method also explains alot
– siddhartha pachhai
8 mins ago




Thanks, this method also explains alot
– siddhartha pachhai
8 mins ago

















 

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