It is estimated that 0.5\% of the population of a city has certa

Monincbh 2021-11-12 Answered
It is estimated that \(\displaystyle{0.5}\%\) of the population of a city has certain disease. A diagnostic test has a probability 0.95 of giving a positive result when applied to a person has the disease, and a probability 0.10 of giving a false positive result. Suppose that the test is now done to a person from the city.
Calculate the following probabilities:
(a) The probability that the test result will be positive;
(b) The probability that given a positive result, the person has the disaese;
(c) The probability that given a negative result, the person does not have the disease;
(d) The probability that the person will be misclassified

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Expert Answer

Anot1954
Answered 2021-11-13 Author has 780 answers
Step 1
a) Obtain the probability that test will be positive.
The probability that test will be positive is obtained below as follows:
From the information, given that
Let P denote the event that the person has tested with positive result.
Let N denote the event that the person has tested with negative result.
Let D denote the event that the person has disease. That is, \(\displaystyle{P}{\left({D}\right)}={0.005}\).
Let ND denote the event that person do not have disease. \(\displaystyle{P}{\left({N}{D}\right)}={0.995}{\left(={1}–{0.005}\right)}\)
Also given that
\(\displaystyle{P}{\left({P}{\mid}{D}\right)}={0.95}\)
\(\displaystyle{\left\lbrace{\left({N}{\mid}{D}\right)}={0.05}{\left(={1}-{0.95}\right)}\right.}\)
\(\displaystyle{P}{\left({P}{\mid}{N}{D}\right)}={0.10}\)
\(\displaystyle{P}{\left({N}{\mid}{N}{D}\right)}={0.90}{\left({1}-{0.10}\right)}\)
The required probabiluty is,
\(\displaystyle{P}{\left({P}\right)}={\left[{P}{\left({P}\cap{D}\right)}\right]}+{\left[{P}{\left({P}\cap{N}{D}\right)}\right]}\)
\(\displaystyle={\left[{P}{\left({P}{\mid}{D}\right)}\times{P}{\left({D}\right)}\right]}+{\left[{P}{\left({P}{\mid}{N}{D}\right)}\times{P}{\left({N}{D}\right)}\right]}\)
\(\displaystyle={\left[{0.95}\times{0.005}\right]}+{\left[{0.10}\times{0.995}\right]}\)
\(\displaystyle={0.00475}+{0.0995}\)
\(\displaystyle={0.10425}\)
\(\displaystyle\approx{0.1043}\)
The probability that test will be positive is 0.1043.
Step 2
b) Obtain the probability that given a positive result, the person has the disease.
The probability that given a positive result, the person has the disease is obtained below as follows:
The required probability is,
\(\displaystyle{P}{\left({D}{\mid}{P}\right)}={\frac{{{P}{\left({P}{\mid}{D}\right)}\times{P}{\left({D}\right)}}}{{{P}{\left({P}\right)}}}}{\left[\therefore{P}{\left({D}\cap{P}\right)}={P}{\left({P}{\mid}{D}\right)}\times{P}{\left({D}\right)}\right]}\)
\(\displaystyle={\frac{{{0.95}\times{0.005}}}{{{0.10425}}}}\)
\(\displaystyle={\frac{{{0.00475}}}{{{0.10425}}}}\)
\(\displaystyle\approx{0.0456}\)
The probability that given a positive result, the person has the disease is 0.0456.
Step 3
c ) Obtain the probability that given a negative result, the person does not have the disease.
The probability that given a negative result, the person does not have the disease is obtained below as follows:
The required probability is,
\(\displaystyle{P}{\left({N}{D}{\mid}{N}\right)}={\frac{{{P}{\left({N}{\mid}{N}{D}\right)}\times{P}{\left({N}{D}\right)}}}{{{P}{\left({N}\right)}}}}{\left[\therefore{P}{\left({N}\cap{N}{D}\right)}={P}{\left({N}{\mid}{N}{D}\right)}\times{P}{\left({N}{D}\right)}\right]}\)
\(\displaystyle={\frac{{{0.90}\times{0.995}}}{{{1}-{0.10425}}}}{\left[{P}{\left({N}\right)}={1}-{P}{\left({P}\right)}\right]}\)
\(\displaystyle={\frac{{{0.8955}}}{{{0.89575}}}}\)
\(\displaystyle={0.9997}\)
The probability that given a negative result, the person does not have the disease is 0.9997.
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