Respuesta :
Answer:
8% probability that he or she actually has the disease
Step-by-step explanation:
We use the Bayes Theorem to solve this question.
Bayes Theorem:
Two events, A and B.
[tex]P(B|A) = \frac{P(B)*P(A|B)}{P(A)}[/tex]
In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.
If a randomly chosen person is given the test and the test comes back positive for conditionitis, what is the probability that he or she actually has the disease?
This means that:
Event A: Test comes back positive.
Event B: Having the disease.
Test coming back positive:
2% have the disease(meaning that P(B) = 0.02), and for those, the test comes positive 98% of the time. This means that [tex]P(A|B) = 0.98[/tex]
For the 100-2 = 98% who do not have the disease, the test comes back positive 100-77 = 23% of the time.
Then
[tex]P(A) = 0.02*0.98 + 0.98*0.23 = 0.245[/tex]
Finally:
[tex]P(B|A) = \frac{0.02*0.98}{0.245} = 0.08[/tex]
8% probability that he or she actually has the disease
