Could some kind people give me the detailed solution of how to solve the below S
ID: 3203966 • Letter: C
Question
Could some kind people give me the detailed solution of how to solve the below Stats question? Thank you! I am trying my best to learn.
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.
(a) What is the joint probability function of patients' health and test
results?
(b) If the test is positive, what is the probability that the patient is
actually sick?
(c) The probability you just calculated in part 1b is the (?)
probability of (?) given (?) .
Explanation / Answer
here let probability of giving a positive =P(+ve) and negative =P(-ve)
, probabilty 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 probabilty of being sick and test positve =P(S)*P(+ve|S) =0.15*0.9 =0.135
and probabilty of being sick and test negative =P(S)*P(-ve|S) =0.15*0.1 =0.015
probabilty of being healthy and test posive =P(H)*P(+veH) =0.85*0.1 =0.085
probabilty of being healthy and test negative =P(H)*P(-ve|H) =0.85*0.9 =0.765
hence joint distribution is as follows:
a)
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 probabilty of being sick given test positve
+ve -ve Total sick 0.135 0.015 0.15 healthy 0.085 0.765 0.85 total 0.22 0.78 1