A Diagnostic test has a 98% probability of giving a positive result when given to a person who has a certain disease. It has a 10% probabili

Question

A Diagnostic test has a 98% probability of giving a positive result when given to a person who has a certain disease. It has a 10% probability of giving a (false) positive result when given to a person who does not have the disease. It is estimated that 15% of the population suffers from this disease. (a) What is the probability that a test result is positive? (b) A person receives a positive test result. What is the probability that this person actually has the discase? (c) A person receives a positive test result. What is the probability that this person does not actually have the disease?

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Eliza 1 week 2022-01-08T04:24:32+00:00 1 Answer 0 views 0

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    2022-01-08T04:26:09+00:00

    Answer:

    a)  P(+) = 0.98*0.15 + 0.1*0.85 = 0.232

    b)  P(+ \cap D)= P(+|D) *P(D)= 0.98*0.15=0.147

     P(D|+) =\frac{0.147}{0.232}=0.634

    c)  P(+ \cap ND)= P(+|ND) *P(ND)= 0.1*0.85=0.085

     P(ND|+) = \frac{0.085}{0.232}= 0.366

    Step-by-step explanation:

    For this case we define the following notation:

    + represent the event of getting a positive result

    D = represent the event of having the disease

    ND = represent the event of NO having the disease

    From the info given we know that:

     P(+|D) = 0.98 , P(+|ND) = 0.1 , P(D) = 0.15

    And by the complement rule we can find:

     P(ND) = 1-P(D) = 1-0.15=0.85

    Part a

    For this case we want to find this probability P(+) and for this case we can use the Bayes total rule and we can do this:

     P(+) = P(+|D) P(D) + P(+|ND) P(ND)

    And if we replace we got:

     P(+) = 0.98*0.15 + 0.1*0.85 = 0.232

    Part b

    We want to find this probability  P(D|+) and using the bayes rule we have:

     P(D|+) = \frac{P(+\ cap D)}{P(+)}

    We can find the numerator from:

     P(+|D) = \frac{P(+ \cap D)}{P(D)}

     P(+ \cap D)= P(+|D) *P(D)= 0.98*0.15=0.147

    And then if we replace we got:

     P(D|+) =\frac{0.147}{0.232}=0.634

    Part c

    We want to find this probability  P(ND|+) and using the bayes rule we have:

     P(ND|+) = \frac{P(+\ cap ND)}{P(+)}

    We can find the numerator from:

     P(+|ND) = \frac{P(+ \cap ND)}{P(ND)}

     P(+ \cap ND)= P(+|ND) *P(ND)= 0.1*0.85=0.085

    And then if we replace we got:

     P(ND|+) = \frac{0.085}{0.232}= 0.366

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