Answer:
The probability that they have the disease given that the test came back positive is 0.6899.
Explanation:
Let A represent the event that the prevalence of a certain disease is 0.09
Let D be the event in which those patients are tested who already have the disease then the probability of B
P (D) = 0.9
Let F be the event in which those patients are tested who do not have the disease then the probability of F
P (F) = 0.96
The tree diagram helps explain the events and their probabilities.
A 0.09 -----------0.9 (D)= P(D/A) (positive test)
⇅------------0.1 ( negative test)
⇅
1----- ⇅
⇅
⇅
B 0.91 ------------0.96 (F)= P(F/ B) (negative test)
----------------0.04 P (G/B) (positive test)
By Bayes Theorem
P (A/ D)= P (A). P(D/A)/ P (A). P(D/A)+P (B) P(G/ B)
P (A/ D)= 0.09 (0.9)/ 0.09 (0.9)+ (0.91)(0.04)
P (A/ D)= 0.081/ 0.081+0.0364
P (A/ D)= 0.081/ 0.1174
P (A/ D)= 0.6899
The probability that they have the disease given that the test came back positive is 0.6899.