Complete Question
Suppose there was a cancer diagnostic test was 95% accurate both on those that do and 90% on those do not have the disease. If 0.4% of the population have cancer, compute the probability that a particular individual has cancer, given that the test indicates he or she has cancer.
Answer:
The probability is [tex]P(C | A) = 0.0042[/tex]
Step-by-step explanation:
From the question we are told that
The probability that the test was accurate given that the person has cancer is
[tex]P(A | C) = 0.95[/tex]
The probability that the test was accurate given that the person do not have cancer is
[tex]P(A | C') = 0.90[/tex]
The probability that a person has cancer is
[tex]P(C) = 0.004[/tex]
Generally the probability that a person do not have cancer is
[tex]P(C') = 1- P(C)[/tex]
=> [tex]P(C') = 1- 0.004[/tex]
=> [tex]P(C') = 0.996[/tex]
Generally the probability that a particular individual has cancer, given that the test indicates he or she has cancer is according to Bayes's theorem evaluated as
[tex]P(C | A) = \frac{P(A | C) * P(C)}{P(A|C) * P(C) + P(A| C') * P(C')}[/tex]
=> [tex]P(C | A) = \frac{ 0.95 * 0.004 }{ 0.95 * 0.004 + 0.90 * 0.996}[/tex]
=> [tex]P(C | A) = 0.0042[/tex]
