Respuesta :
Answer:
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:
[tex]P(A_{n}|X)=\frac{P(X|A_{n})P(A_{n})}{\sum\limits^{n}_{i=1}{P(X|A_{i})P(A_{i})}}[/tex]
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:
[tex]P(A)=0.013\\P(X|A)=0.99\\P(X^{c}|A^{c})=0.89[/tex]
The probability that the test result is positive given that the person does not have HIV\AIDS is:
[tex]P(X|A^{c})=1-P(X^{c}|A^{c})=1-0.89=0.11[/tex]
Compute the value of P (A|X) using the Bayes' theorem as follows:
[tex]P(A|X)=\frac{P(X|A)P(A)}{P(X|A)P(A)+P(X|A^{c})P(A^{c})}[/tex]
[tex]=\frac{(0.99\times 0.013)}{(0.99\times 0.013)+(0.11\times (1-0.013))}[/tex]
[tex]=\frac{0.01287}{0.01287+0.10857}[/tex]
[tex]=0.105978\\\approx0.1060[/tex]
Thus, the probability that a person has HIV\AIDS given that the person was tested positive for HOV|AIDS is 0.1060.
