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Trouble understanding Naive 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?
probability naive-bayes
asked 1 hour ago
siddhartha pachhai
414
add a comment |Â
up vote
1
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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?
probability naive-bayes
asked 1 hour ago
siddhartha pachhai
414
The final+in the denominator should bex.
– BruceET
50 mins ago
1
There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
48 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was watching a video on YouTube and i am not sure if the given solution is correct. Can someone confirm?
probability naive-bayes
asked 1 hour ago
siddhartha pachhai
414
I was watching a video on YouTube and i am not sure if the given solution is correct. Can someone confirm?
probability naive-bayes
probability naive-bayes
asked 1 hour ago
siddhartha pachhai
414
asked 1 hour ago
siddhartha pachhai
414
asked 1 hour ago
siddhartha pachhai
414
asked 1 hour ago
siddhartha pachhai
414
asked 1 hour ago
siddhartha pachhai
414
414
The final+in the denominator should bex.
– BruceET
50 mins ago
1
There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
48 mins ago
add a comment |Â
The final+in the denominator should bex.
– BruceET
50 mins ago
1
There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
48 mins ago
The final
+ in the denominator should be x.– BruceET
50 mins ago
The final
+ in the denominator should be x.– BruceET
50 mins ago
1
1
There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
48 mins ago
There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
48 mins ago
add a comment |Â
2 Answers
2
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up vote
1
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Yes even without consulting the equations it is possible to work it out from the information. See below.
answered 1 hour ago
Curtis White
1112
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.
add a comment |Â
up vote
1
down vote
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'):
$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD^c).$$
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."
answered 1 hour ago
BruceET
3,7031519
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Yes even without consulting the equations it is possible to work it out from the information. See below.
answered 1 hour ago
Curtis White
1112
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.
add a comment |Â
up vote
1
down vote
Yes even without consulting the equations it is possible to work it out from the information. See below.
answered 1 hour ago
Curtis White
1112
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.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes even without consulting the equations it is possible to work it out from the information. See below.
answered 1 hour ago
Curtis White
1112
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.
answered 1 hour ago
Curtis White
1112
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 1 hour ago
Curtis White
1112
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 1 hour ago
Curtis White
1112
answered 1 hour ago
Curtis White
1112
1112
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.
add a comment |Â
add a comment |Â
up vote
1
down vote
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'):
$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD^c).$$
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."
answered 1 hour ago
BruceET
3,7031519
add a comment |Â
up vote
1
down vote
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'):
$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD^c).$$
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."
answered 1 hour ago
BruceET
3,7031519
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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'):
$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD^c).$$
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."
answered 1 hour ago
BruceET
3,7031519
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'):
$$P(D|T) = fracP(DT)P(T) =
fracP(D)P(TP(DT)+P(D^cT)
= fracP(D)P(TD^c).$$
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."
answered 1 hour ago
BruceET
3,7031519
edited 49 mins ago
answered 1 hour ago
BruceET
3,7031519
answered 1 hour ago
BruceET
3,7031519
answered 1 hour ago
BruceET
3,7031519
3,7031519
add a comment |Â
add a comment |Â
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The final
+in the denominator should bex.– BruceET
50 mins ago
1
There is no such a thing as "naive Bayes theorem", there are naive Bayes algorithm and Bayes theorem.
– Tim♦
48 mins ago