Answer:
P (A ║ B) = 1.98 %
Step-by-step explanation:
Bayes´ Theorem express
P (A ║ B) = P(A) * P( B ║ A) / P(B)
Now we identify
Event A person infected with a virus. Probability of being infected by a virus is P/A) 0.001
Event B the test was positive. Probability of test positive P(B) = 0,05
Probability of P ( B║ A) is the porbability of test positive given that is infected = 0.99
Then by subtitution in a general equation of the theorem we have
P (A ║ B) = 0.001*0.99/ 0.05
P (A ║ B) = 0.0198 P (A ║ B) = 1.98 %
Answer:
0.0194
Step-by-step explanation:
Let's define the following events
A: a person is infected with the virus
B: a person test positive
[tex]\bar{A}[/tex]: a person is not infected with the virus
We know that P(A) = 0.001, and then [tex]P(\bar{A})[/tex] = 0.999. A test used to detect the virus is positive 99% of the time if the person test has the virus is equivalent to P(B|A) = 0.99. And positive 5% of the time if the person does not have the virus means [tex]P(B|\bar{A})[/tex]=0.05. We want to find the probability that a person is infected with the virus, given that they test positive, i.e., P(A|B). Bayes Theorem tell us that [tex]P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\bar{A})P(\bar{A})}=\frac{(0.99)(0.001)}{(0.99)(0.001)+(0.05)(0.999)}[/tex]=0.0194
