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.

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.