Step 1

For the given problem let us assume ,

D = event that people have diabetes

N = event that people have no diabetes

C = event that people diagnosed having diabetes

F = event that people diagnosed not having diabetes

Step 2

Now from the given information in the problem, we calculate the probabilities as shown,

\(\displaystyle{P}{\left({D}\right)}={0.0836}\)

\(\displaystyle\Rightarrow{P}{\left({N}\right)}={1}-{0.0836}={0.9164}\)

\(\displaystyle\text{Also},{P}{\left({C} \mid {D}\right)}={0.945}\)

\(\displaystyle\Rightarrow{P}{\left({F}\mid{D}\right)}={1}-{0.945}={0.055}\)

\(\displaystyle\text{And},{P}{\left({C}\mid{N}\right)}={0.02}\)

\(\displaystyle\Rightarrow{P}{\left({F}\mid{N}\right)}={1}-{0.02}={0.98}\)

Step 3

1. The probability that a randomly selected adult over 40 doesn't have diabetes and is diagnosed as having diabetes (such diagnoses are called "false positives") is P(N and C) and is calculated as shown,

\(\displaystyle{P}{\left({N}\ \text{and}\ {C}\right)}={P}{\left({N}\right)}\times{P}{\left({C}\mid{N}\right)}\)

\(\displaystyle={0.9164}\times{0.02}\)

\(\displaystyle={0.018328}\)

\(\displaystyle\approx{0.018}\)

Step 4

Answer. 1: The probability that a randomly selected adult over 40 doesn't have diabetes and is diagnosed as having diabetes (such diagnoses are called "false positives") is 0.018

Step 5

2. The probability that a randomly selected adult of 40 is diagnosed as not having diabetes is P(F) and is calculated as shown,

\(\displaystyle{P}{\left({F}\right)}={P}{\left({F}\ \text{and}\ {D}\right)}+{P}{\left({F}\ \text{and}\ {N}\right)}\)

\(\displaystyle={P}{\left({F}\mid{D}\right)}\times{P}{\left({D}\right)}+{P}{\left({F}\mid{N}\right)}\times{P}{\left({N}\right)}\)

\(\displaystyle={0.055}\times{0.0836}+{0.98}\times{0.9164}\)

\(\displaystyle={0.004598}+{0.898072}\)

\(\displaystyle={0.90267}\approx{0.9027}\)

Step 6

Answer.2: The probability that a randomly selected adult of 40 is diagnosed as not having diabetes is 0.9027

Step 7

3. The probability that a randomly selected adult over 40 actually has diabetes, given that he/she is diagnosed as not having diabetes (such diagnoses are called "false negatives") is \(\displaystyle{P}{\left({D}\mid{F}\right)}\) and is calculated as shown,

\(\displaystyle{P}{\left({D}\mid{F}\right)}={\frac{{{P}{\left({F}\ \text{and}\ {D}\right)}}}{{{P}{\left({F}\right)}}}}\)

\(\displaystyle={\frac{{{P}{\left({F}\mid{D}\right)}\times{P}{\left({D}\right)}}}{{{P}{\left({F}\right)}}}}\)

\(\displaystyle={\frac{{{0.055}\times{0.0836}}}{{{0.9027}}}}\)

\(\displaystyle={0.00509}\approx{0.0051}\)

Step 8

Answer. 3: The probability that a randomly selected adult over 40 actually has diabetes, given that he/she is diagnosed as not having diabetes (such diagnoses are called "false negatives") is 0.0051.