Answer:
0.0893
Step-by-step explanation:
Let A be the event that the individual has the disease
Let Aⁿ be the event that the individual doesn't have the disease
Let B be the event of a positive result showing.
Now,
We are told One in 1000 adults is afflicted with the disease.
Thus;
P(A) = 1/1000 = 0.001
P(Aⁿ) = 1 - 0.001 = 0.999
Also,we are told that when an individual has the disease, a positive result will occur 98% of the time.
Thus;
P(A|B) = 98% = 0.98
Also, we are told that an individual without the disease will show a positive result only 1% of the time. Thus;
P(Aⁿ|B) = 1% = 0.01
Now, from fundamental rule, let's find P(B). Thus;
P(B) = [P(A|B) × P(A)] + [P(Aⁿ|B) × P(Aⁿ)]
P(B) = (0.98 × 0.001) + (0.01 × 0.999)
P(B) = 0.01097
Now, from Baye's theorem, If a randomly selected individual is tested and the result is positive, the probability that the individual has the disease is given by;
P(A|B) = [P(A|B) × P(A)]/P(B)
P(A|B) = (0.98 × 0.001)/0.01097
P(A|B) = 0.0893
