One in 1000 adults is afflicted with a rare disease forwhich a diagnostic test has been developed. The test is such, that when an individual has the disease, a positive result willoccur 98% of the time, while an individual without the disease willshow a positive result only 1% of the time. If a randomlyselected individual is tested and the result is positive, what isthe probability that the individual has the disease?

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

i think it is 70

step-by-step explanation:

8

step-by-step explanation: