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HWKsolutions3lastsemester

# Whatistheprobabilitythattheperson doeshavethedisease

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Unformatted text preview: lly does not have the disease (i.e., 91% specificity). Suppose that 14% of the population has the disease. Note: MAKE A TREE P(Per = + ) = 0.14 P(Per = ‐ ) = .86 P( T = + | Per = + ) = 0.82 P(T = ‐ | Per = +) = .18 P( T = ‐ | Per = ‐ ) = 0.91 P(T = + | Per = ‐ ) = 0.09 o What is the probability that a randomly chosen person will test positive? ANS) P(Test = + ) = P(Test = + |Per = + )*P(Per = +) + P(Test = + | Per = ‐ ) * P(Per = ‐ ) = (.82)(.14) + (.09)(.86) = 0.1922 o Suppose that a randomly chosen person does test positive. What is the probability that the person does have the disease? ANS) “Invert the conditioning” then use “Bayes Theorem” x / (x+y) to compute denom. P (Per = +| T = + ) = P ( T = + | Per = +) * P(Per = +) / P ( T = +) = Denominator: P(T = +) = [P ( T = + | Per = +)P(Per = +) + P ( T = +| Per = ‐ ) P(Per = ‐) P ( Per = + | T = +) = (0.82)(0.14) / [(0.82)(0.14) + (0.09)(0.86)] = 0.5972945 2 http://www.stat.ucla.edu/%7Edinov/courses_students.dir/12/Winter/STAT13.1.dir/HWs.dir/HW03.html (HW3.3) In a certain population of the European starli...
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