Question

Suppose a test for diagnosing a certain serious disease is successful in detecting the disease in 95% of all persons infected, but that it incorrectly diagnoses 4% of all healthy people as having the serious disease. Suppose also that it incorrectly diagnoses 12% of all people having another minor disease as having the serious disease. If it is known that 2% of the population has the serious, 90% of the population is health and 8% has the minor disease, find the probability that a person selected at random has the serious disease if the test indicates that he or she does. Use H to represent healthy, M to represent having minor disease and D to represent having serious disease. Show the tree diagram and contingency table.

Answer 1

Answer:

Step-by-step explanation:

Hello!

In the study population, there is an incidence of two diseases, a serious disease (D) and a minor disease (M).

There is a diagnostic test for the serious disease that successfully detects the disease (positive) in 95% of all persons infected (D). Symbolized P(+/D)=0.95

This test incorrectly diagnoses 4% (positive) of all healthy people (H) as having a serious disease. Symbolized P(+/H)= 0.04

And it incorrectly diagnoses 12% of all people having another minor disease as having a serious disease. Symbolized P(+/M)= 0.12

It is known in the population that:

2% of the population has a serious disease. P(D)= 0.02

90% of the population is healthy. P(H)= 0.90

8% of the population has a minor disease. P(M)= 0.08

I've attached a contingency table with the symbolized probabilities you need to calculate for each category.

Remember: Given the dependent events A and B a conditional probability of this events is defined as P(A/B)= P(A∩B)/P(B)

⇒ Then you can clear the probability of the intersection of both events as:

P(A∩B)= P(A/B)*P(B)

You can apply this to the asked probabilities:

P(+∩D)= P(+/D)*P(D)= 0.95*0.02= 0.019

P(+∩M)= P(+/M)*P(M)= 0.12*0.08= 0.0096

P(+∩H)= P(+/H)*P(H)= 0.04*0.90= 0.036

P(+) = P(+∩D) + P(+∩M) + P(+∩H)= 0.019 + 0.0096 + 0.036= 0.0646

P(-) = 1 - P(+) = 1 - 0.0646= 0.9354

P(-∩D)= P(D) - P(+∩D)= 0.02 - 0.019= 0.001

P(-∩M)= P(M) - P(+∩M)= 0.08 - 0.0096=0.0704

P(-∩H)= P(H) - P(+∩H)= 0.90 - 0.036= 0.864

Tree diagram and completed table attached below.

I hope it helps!