Mathematics: 2.4.31. medical records show that 0.01% of the general adult population not belonging to a high-risk group (for example, intravenous drug users) are hiv-positive. blood tests for the virus are 99.9% accurate when given to someone infected and 99.99% accurate when given to someone not infected. what is the probability that a random adult not in a highrisk group will test positive for the hiv virus? larsen, richard j.. an introduction to mathematical statistics and its applications (p. 44). pearson

Answer :its 11 Step-by-step explanation: cause my dog told me the answer and he/ she always right UwU...

2.4.31. medical records show that 0.01% of the general

sherlock19

05.01.2022

The probability is 0.00019989

Step-by-step explanation:

We are going to define the conditional probability as :

$P(A/B)=\frac{P(A,B)}{P(B)} \\P(B) 0$

Where P(A,B) is the probability of the event (A∩B).

Let's define the following events :

B : ''The HIV test is positive''

A : ''The adult is infected''

Also we define :

$P(A)=1-P(A^{c} )$

Where $A^{c}$ is the event where A does not occur.

In the exercise :

$P(A)=0.0001\\P(B/A)=0.999\\P(B^{c} /A^{c} )=0.9999$

Where

$P(B/A)\\P(B^{c}/A^{c})$

represents the test accuracy

We are ask about P(B).

We write :

$P(A^{c})=1-P(A)=1-0.0001=0.9999\\ P(A^{c})=0.9999$

$P(B/A)=\frac{P(B,A)}{P(A)} \\P(B,A)=P(B/A).P(A)$

$P(B/A^{c})=1-P(B^{c}/A^{c})=1-0.9999=0.0001\\ P(B/A^{c})=0.0001$

$P(B/A^{c})=\frac{P(B,A^{c})}{P(A^{c})} \\$

$P(B,A^{c})=P(B/A^{c}).P(A^{c})$

P(B) = P{ [(B∩A)∪[B∩(A^{c})]}

P(B) = P(B∩A) +P[B∩(A^{c})]

Using the conditional equations

$P(B)=P(B/A).P(A)+P(B/A^{c} ).P(A^{c})\\ P(B)=(0.999).(0.0001)+(0.0001).(0.9999)=(1.9989).10^{-4} =0.00019989\\P(B)=0.00019989$

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