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22savage2017

22.07.2022

Mathematics

The probability that a person has HIV\AIDS given that the person was tested positive for HOV|AIDS is 0.1060.

Step-by-step explanation:

Let a set be events that have occurred be denoted as:

S = {A₁, A₂, A₃,..., Aₙ}

The Bayes' theorem states that the conditional probability of an event, say Aₙ given that another event, say X has already occurred is given by:

$P(A_{n}|X)=\frac{P(X|A_{n})P(A_{n})}{\sum\limits^{n}_{i=1}{P(X|A_{i})P(A_{i})}}$

The screening test for HIV/AIDS designed by a pharmaceutical company is being analysed.

Denote the events as follows:

A = a person has HIV/AIDS

X = the test turns out as positive

The information provided is:

$P(A)=0.013\\P(X|A)=0.99\\P(X^{c}|A^{c})=0.89$

The probability that the test result is positive given that the person does not have HIV\AIDS is:

$P(X|A^{c})=1-P(X^{c}|A^{c})=1-0.89=0.11$

Compute the value of P (A|X) using the Bayes' theorem as follows:

$P(A|X)=\frac{P(X|A)P(A)}{P(X|A)P(A)+P(X|A^{c})P(A^{c})}$

$=\frac{(0.99\times 0.013)}{(0.99\times 0.013)+(0.11\times (1-0.013))}$

$=\frac{0.01287}{0.01287+0.10857}$

$=0.105978\\\approx0.1060$

Thus, the probability that a person has HIV\AIDS given that the person was tested positive for HOV|AIDS is 0.1060.

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