Answer to Question #329957 in Statistics and Probability for khan

Question

A diagnostic test has a probability 0.95 of giving a positive result when applied to a person suffering from a certain disease, and a probability0.10 of giving a (false) positive when applied to a non- sufferer. It is estimated that 0.5 % of the population are sufferers. Suppose that the test is now administered to a person about whom we have no relevant information relating to the disease (apart from the fact that he/she comes from this population). Calculate the following probabilities:

(a) that the test result will be positive.

(b) that, given a positive result, the person is a sufferer.

(c) that, given a negative result, the person is a non-sufferer.

Answer 1

Let A - the person is suffering from a disease,

B - the test is positive,

Then $P(A)=0.005,~P(B|A)=0.95,~P(B|\lnot A)=0.1$

a) We can split probability of event B into two parts: probability of B when A occurred + probability of B when A did not occur:

$P(B)=P(B|A)P(A)+P(B|\lnot A)P(\lnot A)=\\ 0.95\cdot0.005+0.1\cdot0.995=0.10425\approx10.4\%$

b) According to the Bayes' theorem:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}=\\ =\frac{0.95\cdot0.005}{0.10425}\approx0.0456=4.56\%$

c) According to the Bayes' theorem:

$P(\lnot A|\lnot B)=\frac{P(\lnot B|\lnot A)P(\lnot A)}{P(\lnot B)}$

$P(A)+P(\lnot A)=1,$ because events $A$ and $\lnot A$ are complementary

Thus:

$P(\lnot A)=1-P(A)=1-0.005=0.995\\ P(\lnot B)=1-P(B)=1-0.10425=0.89575$

Given that the event $\lnot A$ happened, events $B$ and $\lnot B$ remain complementary:

$P(B|\lnot A)+P(\lnot B|\lnot A)=1\Rarr\\ \Rarr P(\lnot B|\lnot A)=1-P(B|\lnot A)=\\ =1-0.1=0.9$

$P(\lnot A|\lnot B)=\frac{P(\lnot B|\lnot A)P(\lnot A)}{P(\lnot B)}=\\ =\frac{0.9\cdot0.995}{0.89575}\approx0.9997=99.97\%$