Question

1. A lab test produces a positive result with 90% probability when the patient is actually sick and with 10% if the patient is healthy. It is known that 15% of the population is sick.

   1. (a) What is the joint probability function of patients’ health and test results?

   2. (b) If the test is positive, what is the probability that the patient is actually sick?

   3. (c) The probability you just calculated in part 1b is the probability of given .

Answer 1

Answer:

here let probability of giving a positive =P(+ve) and negative =P(-ve)

, probability of being sick =P(S)=0.15, and being healthy =P(H)=1-P(S) =1-0.15=0.85

from given data P(+ve|S) =0.9

P(+ve|H) =0.10

a) hence probability of being sick and test positve =P(S)*P(+ve|S) =0.15*0.9 =0.135

and probability of being sick and test negative =P(S)*P(-ve|S) =0.15*0.1 =0.015

probability of being healthy and test posive =P(H)*P(+veH) =0.85*0.1 =0.085

probability of being healthy and test negative =P(H)*P(-veH) =0.85*0.9 =0.765

hence joint distribution is as follows:

a)

	+Ve	_Ve	Total
sick	0.135	0.015	0.15
healthy	0.085	0.765	0.85
total	0.22	0.78	1

b) probability that the patient is
actually sick, given test is +ve =P(S|T) =0.135/0.22 =0.6136

c)probability you just calculated in part 1b is the probability of being sick given test positve