What's the probability that person is healthy?Bayes Rule questionProbability Related QuestionWhat's the probability 2 people sharing a house will get the same disease?Conditional probability -exerciseConditional probability related geneticsConditional probability question; Bayes' theory and medical testProbability of an event given a certain conditionProbability question: what's the sample space?What is the probability that the person actually has the disease?Disease test - probability for a person with a positive result to be healthy
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What's the probability that person is healthy?
Bayes Rule questionProbability Related QuestionWhat's the probability 2 people sharing a house will get the same disease?Conditional probability -exerciseConditional probability related geneticsConditional probability question; Bayes' theory and medical testProbability of an event given a certain conditionProbability question: what's the sample space?What is the probability that the person actually has the disease?Disease test - probability for a person with a positive result to be healthy
$begingroup$
My question looks like this:
Using X-ray probability to find out that person is sick: a=0.98. Probability to accidentally identify the disease for health person: b=0.3. Let's say that there is 10 percent of sick people in the world. What's the probability, that person is healthy even after he was identified a ill person?
My solution:
P(A)= 0.98*0.3*0.1=0.0294
But I'm not sure if this is correct.
Any help appreciate, thanks!
probability probability-theory
asked Mar 21 at 8:22
MCTGMCTG
194
$endgroup$
add a comment |
$begingroup$
My question looks like this:
Using X-ray probability to find out that person is sick: a=0.98. Probability to accidentally identify the disease for health person: b=0.3. Let's say that there is 10 percent of sick people in the world. What's the probability, that person is healthy even after he was identified a ill person?
My solution:
P(A)= 0.98*0.3*0.1=0.0294
But I'm not sure if this is correct.
Any help appreciate, thanks!
probability probability-theory
asked Mar 21 at 8:22
MCTGMCTG
194
$endgroup$
1
$begingroup$
It is not correct. Work out the probability that a person is sick and is diagnosed as sick, and the probability that a person is healthy but diagnosed as sick. Then combine these to find the overall probability of being diagnosed as sick, and thus find the conditional probability
$endgroup$
– Henry
Mar 21 at 8:31
$begingroup$
Then a=0.98 probability has no impact for the answer?
$endgroup$
– MCTG
Mar 21 at 8:36
1
$begingroup$
I would assume $a=0.98$ affects the probability a sick person is diagnosed as sick (though the wording is less than perfect) and so affects the final answer
$endgroup$
– Henry
Mar 21 at 8:42
add a comment |
$begingroup$
My question looks like this:
Using X-ray probability to find out that person is sick: a=0.98. Probability to accidentally identify the disease for health person: b=0.3. Let's say that there is 10 percent of sick people in the world. What's the probability, that person is healthy even after he was identified a ill person?
My solution:
P(A)= 0.98*0.3*0.1=0.0294
But I'm not sure if this is correct.
Any help appreciate, thanks!
probability probability-theory
asked Mar 21 at 8:22
MCTGMCTG
194
$endgroup$
My question looks like this:
Using X-ray probability to find out that person is sick: a=0.98. Probability to accidentally identify the disease for health person: b=0.3. Let's say that there is 10 percent of sick people in the world. What's the probability, that person is healthy even after he was identified a ill person?
My solution:
P(A)= 0.98*0.3*0.1=0.0294
But I'm not sure if this is correct.
Any help appreciate, thanks!
probability probability-theory
probability probability-theory
asked Mar 21 at 8:22
MCTGMCTG
194
asked Mar 21 at 8:22
MCTGMCTG
194
asked Mar 21 at 8:22
MCTGMCTG
194
asked Mar 21 at 8:22
MCTGMCTG
194
asked Mar 21 at 8:22
MCTGMCTG
194
194
1
$begingroup$
It is not correct. Work out the probability that a person is sick and is diagnosed as sick, and the probability that a person is healthy but diagnosed as sick. Then combine these to find the overall probability of being diagnosed as sick, and thus find the conditional probability
$endgroup$
– Henry
Mar 21 at 8:31
$begingroup$
Then a=0.98 probability has no impact for the answer?
$endgroup$
– MCTG
Mar 21 at 8:36
1
$begingroup$
I would assume $a=0.98$ affects the probability a sick person is diagnosed as sick (though the wording is less than perfect) and so affects the final answer
$endgroup$
– Henry
Mar 21 at 8:42
add a comment |
1
$begingroup$
It is not correct. Work out the probability that a person is sick and is diagnosed as sick, and the probability that a person is healthy but diagnosed as sick. Then combine these to find the overall probability of being diagnosed as sick, and thus find the conditional probability
$endgroup$
– Henry
Mar 21 at 8:31
$begingroup$
Then a=0.98 probability has no impact for the answer?
$endgroup$
– MCTG
Mar 21 at 8:36
1
$begingroup$
I would assume $a=0.98$ affects the probability a sick person is diagnosed as sick (though the wording is less than perfect) and so affects the final answer
$endgroup$
– Henry
Mar 21 at 8:42
1
1
$begingroup$
It is not correct. Work out the probability that a person is sick and is diagnosed as sick, and the probability that a person is healthy but diagnosed as sick. Then combine these to find the overall probability of being diagnosed as sick, and thus find the conditional probability
$endgroup$
– Henry
Mar 21 at 8:31
$begingroup$
It is not correct. Work out the probability that a person is sick and is diagnosed as sick, and the probability that a person is healthy but diagnosed as sick. Then combine these to find the overall probability of being diagnosed as sick, and thus find the conditional probability
$endgroup$
– Henry
Mar 21 at 8:31
$begingroup$
Then a=0.98 probability has no impact for the answer?
$endgroup$
– MCTG
Mar 21 at 8:36
$begingroup$
Then a=0.98 probability has no impact for the answer?
$endgroup$
– MCTG
Mar 21 at 8:36
1
1
$begingroup$
I would assume $a=0.98$ affects the probability a sick person is diagnosed as sick (though the wording is less than perfect) and so affects the final answer
$endgroup$
– Henry
Mar 21 at 8:42
$begingroup$
I would assume $a=0.98$ affects the probability a sick person is diagnosed as sick (though the wording is less than perfect) and so affects the final answer
$endgroup$
– Henry
Mar 21 at 8:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A person can either be healthy or sick with probabilities 0.9 and 0.1 (say $Pr(H)$ and $Pr(barH)$). It is given that the probability that a healthy person maybe accidentally be misdiagnosed is 0.3, this means that $Pr(Positive|H) = 0.3$. We are asked the probability $Pr(H|Positive)$.
Also given that a person is identified with the disease (irrespective of whether or not he has it) as 0.98. That is,
$$
Pr(Positive) = 0.98
$$
By Bayes' rule,
$$
Pr(H|Positive) = fracPr(PositivePr(Positive)
$$
But we can understand some more properties of the drug by exploring,
$$
Pr(Negative|H) = 1- Pr(Positive|H) = 0.7
$$
$$
Pr(Positive) = Pr(Positive|H) + Pr(Positive|barH)
$$
$$
implies Pr(Positive|barH) = Pr(Positive) - Pr(Positive|H) = 0.78
$$
This also means,
$$
Pr(Negative|barH) = 0.22
$$
This drug clearly has an alarming false positive rate. Undertandable as it seems to classify 98% of the population as sick even though only 10% actually are. Like @drhab has pointed out, this is a classic example of Bayes' rule in action.
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
add a comment |
$begingroup$
Hint:
$Pleft(texthealthymidtextill testedright)Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)$
$Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)+Pleft(textill testedmidtextillright)Pleft(textillright)$
Based on these equalities (do you agree that they are correct?) and your data you can find $Pleft(texthealthymidtextill testedright)$.
Give it a try (and "discover" the so-called rule of Bayes).
answered Mar 21 at 8:38
drhabdrhab
104k545136
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A person can either be healthy or sick with probabilities 0.9 and 0.1 (say $Pr(H)$ and $Pr(barH)$). It is given that the probability that a healthy person maybe accidentally be misdiagnosed is 0.3, this means that $Pr(Positive|H) = 0.3$. We are asked the probability $Pr(H|Positive)$.
Also given that a person is identified with the disease (irrespective of whether or not he has it) as 0.98. That is,
$$
Pr(Positive) = 0.98
$$
By Bayes' rule,
$$
Pr(H|Positive) = fracPr(PositivePr(Positive)
$$
But we can understand some more properties of the drug by exploring,
$$
Pr(Negative|H) = 1- Pr(Positive|H) = 0.7
$$
$$
Pr(Positive) = Pr(Positive|H) + Pr(Positive|barH)
$$
$$
implies Pr(Positive|barH) = Pr(Positive) - Pr(Positive|H) = 0.78
$$
This also means,
$$
Pr(Negative|barH) = 0.22
$$
This drug clearly has an alarming false positive rate. Undertandable as it seems to classify 98% of the population as sick even though only 10% actually are. Like @drhab has pointed out, this is a classic example of Bayes' rule in action.
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
add a comment |
$begingroup$
A person can either be healthy or sick with probabilities 0.9 and 0.1 (say $Pr(H)$ and $Pr(barH)$). It is given that the probability that a healthy person maybe accidentally be misdiagnosed is 0.3, this means that $Pr(Positive|H) = 0.3$. We are asked the probability $Pr(H|Positive)$.
Also given that a person is identified with the disease (irrespective of whether or not he has it) as 0.98. That is,
$$
Pr(Positive) = 0.98
$$
By Bayes' rule,
$$
Pr(H|Positive) = fracPr(PositivePr(Positive)
$$
But we can understand some more properties of the drug by exploring,
$$
Pr(Negative|H) = 1- Pr(Positive|H) = 0.7
$$
$$
Pr(Positive) = Pr(Positive|H) + Pr(Positive|barH)
$$
$$
implies Pr(Positive|barH) = Pr(Positive) - Pr(Positive|H) = 0.78
$$
This also means,
$$
Pr(Negative|barH) = 0.22
$$
This drug clearly has an alarming false positive rate. Undertandable as it seems to classify 98% of the population as sick even though only 10% actually are. Like @drhab has pointed out, this is a classic example of Bayes' rule in action.
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
add a comment |
$begingroup$
A person can either be healthy or sick with probabilities 0.9 and 0.1 (say $Pr(H)$ and $Pr(barH)$). It is given that the probability that a healthy person maybe accidentally be misdiagnosed is 0.3, this means that $Pr(Positive|H) = 0.3$. We are asked the probability $Pr(H|Positive)$.
Also given that a person is identified with the disease (irrespective of whether or not he has it) as 0.98. That is,
$$
Pr(Positive) = 0.98
$$
By Bayes' rule,
$$
Pr(H|Positive) = fracPr(PositivePr(Positive)
$$
But we can understand some more properties of the drug by exploring,
$$
Pr(Negative|H) = 1- Pr(Positive|H) = 0.7
$$
$$
Pr(Positive) = Pr(Positive|H) + Pr(Positive|barH)
$$
$$
implies Pr(Positive|barH) = Pr(Positive) - Pr(Positive|H) = 0.78
$$
This also means,
$$
Pr(Negative|barH) = 0.22
$$
This drug clearly has an alarming false positive rate. Undertandable as it seems to classify 98% of the population as sick even though only 10% actually are. Like @drhab has pointed out, this is a classic example of Bayes' rule in action.
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
$endgroup$
A person can either be healthy or sick with probabilities 0.9 and 0.1 (say $Pr(H)$ and $Pr(barH)$). It is given that the probability that a healthy person maybe accidentally be misdiagnosed is 0.3, this means that $Pr(Positive|H) = 0.3$. We are asked the probability $Pr(H|Positive)$.
Also given that a person is identified with the disease (irrespective of whether or not he has it) as 0.98. That is,
$$
Pr(Positive) = 0.98
$$
By Bayes' rule,
$$
Pr(H|Positive) = fracPr(PositivePr(Positive)
$$
But we can understand some more properties of the drug by exploring,
$$
Pr(Negative|H) = 1- Pr(Positive|H) = 0.7
$$
$$
Pr(Positive) = Pr(Positive|H) + Pr(Positive|barH)
$$
$$
implies Pr(Positive|barH) = Pr(Positive) - Pr(Positive|H) = 0.78
$$
This also means,
$$
Pr(Negative|barH) = 0.22
$$
This drug clearly has an alarming false positive rate. Undertandable as it seems to classify 98% of the population as sick even though only 10% actually are. Like @drhab has pointed out, this is a classic example of Bayes' rule in action.
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
answered Mar 21 at 8:50
Balakrishnan RajanBalakrishnan Rajan
1519
1519
add a comment |
add a comment |
$begingroup$
Hint:
$Pleft(texthealthymidtextill testedright)Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)$
$Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)+Pleft(textill testedmidtextillright)Pleft(textillright)$
Based on these equalities (do you agree that they are correct?) and your data you can find $Pleft(texthealthymidtextill testedright)$.
Give it a try (and "discover" the so-called rule of Bayes).
answered Mar 21 at 8:38
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Hint:
$Pleft(texthealthymidtextill testedright)Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)$
$Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)+Pleft(textill testedmidtextillright)Pleft(textillright)$
Based on these equalities (do you agree that they are correct?) and your data you can find $Pleft(texthealthymidtextill testedright)$.
Give it a try (and "discover" the so-called rule of Bayes).
answered Mar 21 at 8:38
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
Hint:
$Pleft(texthealthymidtextill testedright)Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)$
$Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)+Pleft(textill testedmidtextillright)Pleft(textillright)$
Based on these equalities (do you agree that they are correct?) and your data you can find $Pleft(texthealthymidtextill testedright)$.
Give it a try (and "discover" the so-called rule of Bayes).
answered Mar 21 at 8:38
drhabdrhab
104k545136
$endgroup$
Hint:
$Pleft(texthealthymidtextill testedright)Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)$
$Pleft(textill testedright)=Pleft(textill testedmidtexthealthyright)Pleft(texthealthyright)+Pleft(textill testedmidtextillright)Pleft(textillright)$
Based on these equalities (do you agree that they are correct?) and your data you can find $Pleft(texthealthymidtextill testedright)$.
Give it a try (and "discover" the so-called rule of Bayes).
answered Mar 21 at 8:38
drhabdrhab
104k545136
answered Mar 21 at 8:38
drhabdrhab
104k545136
answered Mar 21 at 8:38
drhabdrhab
104k545136
answered Mar 21 at 8:38
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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$begingroup$
It is not correct. Work out the probability that a person is sick and is diagnosed as sick, and the probability that a person is healthy but diagnosed as sick. Then combine these to find the overall probability of being diagnosed as sick, and thus find the conditional probability
$endgroup$
– Henry
Mar 21 at 8:31
$begingroup$
Then a=0.98 probability has no impact for the answer?
$endgroup$
– MCTG
Mar 21 at 8:36
1
$begingroup$
I would assume $a=0.98$ affects the probability a sick person is diagnosed as sick (though the wording is less than perfect) and so affects the final answer
$endgroup$
– Henry
Mar 21 at 8:42