This is a problem which requires Bayes Theorem Leukemia has

This is a problem which requires Bayes\' Theorem. Leukemia has an incidence in the population of 1 in 10,000. Your biotech firm has developed a test for early stages of leukemia which you wish to have approved by the FDA. Your test is found to have a probability of detecting the disease in 99 % of all leukemia patients tested; however, in testing individuals without leukemia the test shows positive in 2 % of that population. What is the probability that a subject who tests positive will have leukemia?

Solution

Ans:

Since 0.01% of the population is infected, then 99.9% is not infected.

Since  1% is negative.

Similarly, we can conclude that 98% of the tests in not infected people is negative.

What we need to find, probability that a person whose test was positive is infected with the Leukamea or in other words that a person is infected given that the test was positive, can be expressed (using the Bayes\' theorem) as

P(P_I/Tp)= p()p(PI)/p(Tp)

Where P means person is infected and T_P means tests are positive.

p (P_I) is 0.01/100=.0001

p (Tp/P_I)= 99/100=0.99

p (T_P), that is the probability that tests are positive. For this purpose, we can use thelaw of the total probability in the following way:

p (T_P)=p(P_I) . p (Tp/P_I) + p(PI). p (Tp/P_I)= 0.001 *0.99 +0.995*0.02

=0.00099 + 0.0199

= 0.02089

P (P_I/Tp) = 0.001 * 0.99/0.02089

=0.04739

=5.0%