Question

stion 6: Cystic fibrosis is a hereditary lung disease that usually results in premature death. It can e inherited if both parents are carriers of an abnormal gene. The CF gene was identified in 1989. The prevalence of this gene in people of European ancestry is 1 in 25, or 4%. The CF20m is a test that is to detect harmful mutations of the CF gene. The sensitivity of the CF20m test is 0.90, and its specificity is 0.98. A person of European decent is randomly chosen and subjected to a CF20m test (a) Compute the probability that the CF20m test is positive. (b) If the result of the CF20m test is positive, what is the probability this person will have a harmful mutation of the CF gene? (c) Supose the result of the CF20m test is negative for this particular person. Compute the probability that this particular person does not have the harmful mutation of the CF gene.

Answer 1

The truth table below is calculated using

P(A person has disease) = TP+FN = 0.04

Note: TP: P(True Positive), FN = P(False Negative), TN: P(True Negative), FP: P(False Positive)

And working out the TP and TN numbers from the definition of Sensitivity and Specificity as below

Sensitivity	TP/(TP+FN)	0.9
	So, TP = 0.9*(TP+FN) = 0.036
Specificity	TN/(TN+FP)	0.98
	So, TN = 0.98*(TN+FP) = 0.98*0.96 = 0.9408

Truth Table:

		Has CF20m	Does not have CF20m	Total
Test Score	Positive	TP = 0.036	FP = 0.96 - 0.9408 = 0.0192	0.036 + 0.0192 = 0.0552
	Negative	FN = 0.04 - 0.036 = 0.004	TN = 0.9408	0.004 + 0.9408 = 0.9448
		TP+FN = 0.04	TN+FP = 0.96

a) P(CF20m is positive) = TP + FP = 0.0552  (detailed working in truth table above)

b) P(person has a harmful mutation / test result is +ve)

= P(Test result is +ve for a person actually having harmful mutation) / P(Test result is +ve) (using conditional probability)

= TP / (TP+FP) = 0.036 / 0.0552 = 0.6522

c) P(Person doesn't have harmful mutation / test result is -ve)

= P(Test result is -ve for a person not having harmful mutation) / P(Test result is -ve)   (again, using conditional probability)

= TN / (TN+FN) = 0.9408 / 0.9448 = 0.9958

(Pls note that I've used TN/FN etc interchangeably for P(TN)/P(FN) etc above for brevity, but these are essentially the probabilities that we are talking about)