Initial diagnosis of a patient’s illness was potentially flu (F), cold (C), or h
ID: 3204973 • Letter: I
Question
Initial diagnosis of a patient’s illness was potentially flu (F), cold (C), or hay fever (H),
with probabilities 0.1, 0.3, and 0.6 respectively. A lab test was then taken from thepatient. Past tests on similar patients show the following conditional probabilities on thetest result being positive (+), neutral (0), or negative (-):
P(+|F) = 0.8 P(0|F) = 0.1 P(-|F) = 0.1P(+|C) = 0.2 P(0|C) = 0.7 P(-|C) = 0.1P(+|H) = 0.1 P(0|H) = 0.3 P(-|H) = 0.6
Use the Bayes Theorem to estimate the respective probabilities of F, C, and H given thata +, 0, or – test result has been observed.
please explain answer
Explanation / Answer
from the given probabilities first we find the P(+),P(-) and P(0)
now
P(+) =P(F)*P(+|F) +P(C)*P(+|C) +P(H )*P(+|H)
=0.1*0.8 +0.3*0.2+0.6*0.1
=0.08+0.06+0.06 =0.2
P(0)=P(F)*P(0|F) +P(C)*P(0|C) +P(H )*P(0|H)
=0.1*0.1+0.3*0.7 +0.6*0.3
=0.4
P(-) =P(F)*P(-|F) +P(C)*P(-|C) +P(H )*P(-|H)
=0.1*0.1+0.3*0.1+0.6*0.6
=0.4
now
P(F|+) =P(F and +)/P(+) =(P(F)*P(+|F) ) /P(+) =0.1*0.8/0.2 =0.4
P(C|+) =P(C and +)/P(+) =(P(C)*P(+|C) ) /P(+) =(0.3*0.2)/0.2=0.3
P(H|+) =P(H and +)/P(+) =(P(H)*P(+|H) ) /P(+) =(0.6*0.1 )/0.2=0.3
P(F|0) =P(F and 0)/P(0) =(P(F)*P(0|F) ) /P(0) =(0.1*0.1)/0.4 =0.025
P(C|0) =P(C and 0)/P(0) =(P(C)*P(0|C) ) /P(0)=(0.3*0.7)/0.4 =0.525
P(H|0) =P(H and 0)/P(0) =(P(H)*P(0|H) ) /P(0)=(0.6*0.3)/0.4 =0.45
P(F|-) =P(F and -)/P(-) =(P(F)*P(-|F) ) /P(-)=(0.1*0.1)/0.4 =0.025
P(C|-) =P(C and -)/P(-) =(P(C)*P(-|C) ) /P(-)=(0.3*0.1) /0.4 =0.075
P(H|-) =P(H and -)/P(-) =(P(H)*P(-|H) ) /P(0)=(0.6*0.6)/0.4 =0.90