 # 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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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.