Question

As a result of medical examination, one of the tests revealed a serious illness in a person. This test has a high precision of 99% (the probability of a positive response in the presence of the disease is 99%, the probability of a negative response in the absence of the disease is also 99%). However, the detected disease is quite rare and occurs only in one person per 10,000. Calculate the probability that the person being examined does have an identi ed disease.

Answer 1

Answer:

0.0098

Step-by-step explanation:

Given,

• Probability of test is positive if the person has disease, P(T/D)=0.99

Probability of test is negative if the person has disease,

P(T'/D) = 1-0.99 = 0.01

• Probability of test is negative if the person hasn't disease, P(T'/D')= 0.99

Probability of test is positive if the person hasn't disease,

P(T/D') = 1 - 0.99 = 0.01

• Probability of occurrence of disease, P(D) = 0.0001

Probability of not occurrence of disease,

P(D) = 1 - 0.0001 = 0.9999

Probability that test will be positive either disease is present or not,

P(T) = P(T/D).P(D)+P(T/D').P(D')  

      =0.99 x 0.0001 + 0.01 x 0.9999

      = 0.000099 + 0.009999

      = 0.010098

So, the probability that the person will have disease if the test is positive,

$P(D/T)\ =\ \dfrac{P(T/D).P(D)}{P(T)}$

           $=\ \dfrac{0.99\times 0.0001}{0.010098}$

            = 0.0098

So, the required probability will be 0.0098.