Respuesta :
Answer:
There is approximately 17% chance of a person not having a disease if he or she has tested positive.
Step-by-step explanation:
Denote the events as follows:
D = a person has contracted the disease.
+ = a person tests positive
- = a person tests negative
The information provided is:
[tex]P(D^{c})=0.95\\P(+|D) = 0.98\\P(+|D^{c})=0.01[/tex]
Compute the missing probabilities as follows:
[tex]P(D) = 1- P(D^{c})=1-0.95=0.05\\\\P(-|D)=1-P(+|D)=1-0.98=0.02\\\\P(-|D^{c})=1-P(+|D^{c})=1-0.01=0.99[/tex]
The Bayes' theorem states that the conditional probability of an event, say A provided that another event B has already occurred is:
[tex]P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^{c})P(A^{c})}[/tex]
Compute the probability that a random selected person does not have the infection if he or she has tested positive as follows:
[tex]P(D^{c}|+)=\frac{P(+|D^{c})P(D^{c})}{P(+|D^{c})P(D^{c})+P(+|D)P(D)}[/tex]
[tex]=\frac{(0.01\times 0.95)}{(0.01\times 0.95)+(0.98\times 0.05)}\\\\=\frac{0.0095}{0.0095+0.0475}\\\\=0.1666667\\\\\approx 0.1667[/tex]
So, there is approximately 17% chance of a person not having a disease if he or she has tested positive.
As the false negative rate of the test is 1%, this probability is not unusual considering the huge number of test done.
