Cardiovascular Disease A group from the University of Utah set up an experiment
ID: 3299691 • Letter: C
Question
Cardiovascular Disease A group from the University of Utah set up an experiment to use Bayes' rule to help make clinical diagnoses [14]. In par ticular, a detailed medical history questionnaire and electro- cardiogram were administered to each patient referred to a cardiovascular laboratory and suspected of having congeni- tal heart disease. From the experience of this laboratory and from estimates based on other published data, two sets of probabilities were generated: (1) The unconditional probability of each of several disease states (see prevalence column in Table 3.12) (2) The conditional probability of specific symptoms given specific disease states (see the rest of Table 3.12). hus, the probability that a person has chest pain given that he or she is normal is .05. Similarly, the proportion of per- sons with isolated pulmonary hypertension is.020. A subset of the data is given in Table 3.12. Assume that these diagnoses are the only ones possible and that a patient can have one and only one diagnosis.Explanation / Answer
Solution
Back-up Theory
If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then
Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(1)
P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(2)
If A is made up of k mutually and collectively exhaustive sub-events, A1, A2,..Ak,
P(B) = sum over i = 1 to k of {P(B/Ai) x P(Ai)} ………………………………..(3)
P(A/B) = P(B/A) x { P(A)/P(B)}……………………………..………………….(4)
Now, to work out the solution,
Let A represent the event ‘Normal’ i.e., Y1 under diagnosis and B represent the joint event of having symptoms (X1) age 1-20 years, (X6) repeated respiratory infections, and (X4) easy fatigue.
Then, we want to find: P(A/B).
Vide (4) of Back-up Theory,
P(A/B) = P(B/A) x { P(A)/P(B)}……………………………………………………(5)
Now, given ‘Suppose we assume the probabilities of any set of symptoms are independent given a specific diagnosis’,
P(B/A) = P(X1/A) x P(X6/A) x P(X4/A)
= 0.49 x 0.05 x 0.1 = 0.00245 …………………………………………………………(6)
P(A) = 0.155 [given] …………………………………………………………………..(7)
Vide (3) of Back-up Theory,
P(B) = sum over i = 1 to 7 of {P(B/Yi) x P(Yi)}
= 0.041236 [Excel calculation details are given at the bottom.] ……………………….(8)
(6), (7) and (8) => P(A/B) = 0.00245 x (0.155/0.041236) = 0.009209 ANSWER
[Excel calculation details]
Diagnosis
Prevalence
x1
x6
x4
P(B/Yi) = x1.x6.x4
P(B/Yi)x P(Yi)
y1
0.155
0.49
0.05
0.1
0.00245
0.00038
y2
0.126
0.5
0.4
0.5
0.1
0.0126
y3
0.084
0.55
0.1
0.9
0.0495
0.004158
y4
0.02
0.45
0.1
0.95
0.04275
0.000855
y5
0.098
0.1
0.05
0.7
0.0035
0.000343
y6
0.391
0.7
0.15
0.3
0.0315
0.012317
y7
0.126
0.6
0.2
0.7
0.084
0.010584
Total
1
0.041236
Diagnosis
Prevalence
x1
x6
x4
P(B/Yi) = x1.x6.x4
P(B/Yi)x P(Yi)
y1
0.155
0.49
0.05
0.1
0.00245
0.00038
y2
0.126
0.5
0.4
0.5
0.1
0.0126
y3
0.084
0.55
0.1
0.9
0.0495
0.004158
y4
0.02
0.45
0.1
0.95
0.04275
0.000855
y5
0.098
0.1
0.05
0.7
0.0035
0.000343
y6
0.391
0.7
0.15
0.3
0.0315
0.012317
y7
0.126
0.6
0.2
0.7
0.084
0.010584
Total
1
0.041236