# It is estimated that aproximately 8.36% Americans are afflicted with Diabetes . Suppose that a ceratin diagnostic evaluation for diabetes will correct

It is estimated that aproximately $8.36\mathrm{%}$ Americans are afflicted with Diabetes .
Suppose that a ceratin diagnostic evaluation for diabetes will correctly diagnose $94.5\mathrm{%}$ of all adults over 40 with diabetes as having the disease and incorrectly diagnoses $2\mathrm{%}$ of all adults over 40 without diabetes as having the disease .
1) Find the probability that a randamly selected adult over 40 doesn't have diabetes and is diagnosed as having diabetes ( such diagnoses are called "false positives").
2) Find the probability that a randomly selected adult of 40 is diagnosed as not having diabetes.
3) Find 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").
Note: It will be helpful to first draw an appropriate tree diagram modeling the situation.
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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,
$P\left(D\right)=0.0836$
$⇒P\left(N\right)=1-0.0836=0.9164$
$\text{Also},P\left(C\mid D\right)=0.945$
$⇒P\left(F\mid D\right)=1-0.945=0.055$
$\text{And},P\left(C\mid N\right)=0.02$
$⇒P\left(F\mid N\right)=1-0.02=0.98$
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,

$=0.9164×0.02$
$=0.018328$
$\approx 0.018$
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,

$=P\left(F\mid D\right)×P\left(D\right)+P\left(F\mid N\right)×P\left(N\right)$
$=0.055×0.0836+0.98×0.9164$
$=0.004598+0.898072$
$=0.90267\approx 0.9027$
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 $P\left(D\mid F\right)$ and is calculated as shown,

$=\frac{P\left(F\mid D\right)×P\left(D\right)}{P\left(F\right)}$
$=\frac{0.055×0.0836}{0.9027}$
$=0.00509\approx 0.0051$
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.