A certain test for a particular cancer is known to be 95% accurate. A person submits to the test and the results are positive. Suppose that the person comes from a population of 100,000 where 2000 people suffer from that disease. What can we conclude about the probability that the person under test has that particular cancer?

Mumbai University > Electronics and Telecommunication > Sem5 > Random Signal Analysis

Marks: 5M

Year: May 2015

Total Population n(S) =100000

Let ‘A’ be the event that the person is under test, ‘B1’ be the event that the person has cancer

∴ Probability that a person has the cancer is

$$P(B_1 )=\frac{2000}{100000}=0.02$$

Hence probability that a person does not have the cancer:

$$P(B_2 )=1-P(B_1 )=1-0.02=0.98$$

Probability the test is positive when a person has cancer is:

$$P(\frac{A}{B_1} )=\frac{95}{100}=0.95$$

Probability the test is positive when a person does not have the cancer is:

$$P(\frac{A}{B_2} )=1-0.95=0.05$$

According to total probability theorem

$$P(A)= \sum_{i=1}^nP(B_i ).P(A/B_i ) $$

$$P(A)=P(\frac{A}{B_1}).P{B_1 }+P(\frac{A}{B_2} ).P(B_2)$$

$$P(A)=0.95×0.02+0.05×0.98$$

$$P(A)=0.0672$$

Using Bayes Theorem

$P(\frac{B_i}{A})=\frac{(P{B_i }.P(\frac{A}{B_i }))}{(P(A))}$

$$P(\frac{B_i}{A})=\frac{(P{B_i }.P(\frac{A}{B_i }))}{\sum_{i=1}^nP(B_i).P(\frac{A}{B_i})}$$

$$P(\frac{B_1}{A})=(0.02×0.95)/(0.0672)=0.2827$$