A laboratory blood test is \(99\%\) defecting a certain disease when it is, infact present.However the test also yields a falls positive result for \(0.5\%\) of the healthy person tested(i.e.,if a healthy person is tested,then with probability 0.005,the test will imply he has the diesease).If \(0.1\%\) of the population actually has the disease,what is the probability that a person has the disease given that his test result is positive?
Solution and Explanation
Let \(E_1\)=The person selected is suffering from certain disease,\(E_2\)=The person selected is not sufferinf from certain disease and A=The doctor
diagnoses correctly
Now, \(P(E_1)=0.1\%=0.001, P(E_2)=1-0.001=0.999,\)
\(P(A|E_1)=99\%=\frac{99}{100}=0.99, P(A|E_2)=0.005\%\)
Therefore,by Bayes'theorem,
\(P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}\)
\(=\frac{0.01×0.99}{0.001×0.99+0.999×0.005}\)
\(=\frac{990}{990+4995}\)
\(=\frac{22}{133}\)
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Concepts Used:
Bayes Theorem
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
- Events A and B are mutually exhaustive events.
- P(A) and P(B) are the probabilities of events A and B, respectively.
- P(A|B) is the conditional probability of the happening of event A, given that event B has happened.
- P(B|A) is the conditional probability of the happening of event B, given that event A has already happened.
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.