Question 14613: It has been estimated that eighty percent of people with coronoary disease have chest pain syptom and twenty-five percent of people without the disease have chest pain symptoms. A person visits a doctor with chest pain symptom, what are the probabilities of the following events?
a) the person has coronary heart disease
b)the person does not have coronary disease.
I have just started my module on probability, I know this maybe pretty basic stuff, but would appreciate any help that will make it easier for me to understand the subject,thanks in advance.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! It has been estimated that eighty percent of people with coronoary disease have chest pain syptom and twenty-five percent of people without the disease have chest pain symptoms. A person visits a doctor with chest pain symptom, what are the probabilities of the following events?
a) the person has coronary heart disease
Looks like a Bayes Problem:
Let "D" mean "has coronary disease"; Let "p" means "has pain".
P(D|P) = P(D and P)/[P(P and D) + P(P and not D)
= 0.80/(0.80 + 0.25) = 0.8/1.05 = 0.762
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b)the person does not have coronary disease.
Also, looks like a Bayes Problem.
P(not D |Pain) = P(not D and Pain)/[P(not D and Pain) + P(not D and not Pain)
=0.25/[0.25+0.20) = 0.556
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Cheers,
Stan H.
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