Question

Suppose 1 in 25 adults is afflicted with a disease for which a new diagnostic test has been developed. Given that an individual actually has the disease, a positive test result will occur with probability .99. Given that an individual does not have the disease, a negative test result will occur with probability .98.

Use a probability tree to answer the following questions.

Question 1. What is the probability of a positive test rest result?

Question 2. If a randomly selected adult is tested and the result is positive, what is the probability that the individual has the disease?

Question 3. If a randomly selected adult is tested and the result is negative, what is the probability that the individual does not have the disease?

Answer 1

Probability tree based on the given data:

Let D = Event that the adult is diseased P= Event that the test is positive H =  Event that the adult is not diseased i.e. healthy N =  Event that the test is negative

1. Probability of a positive test rest result: From the probability tree,

Pr (P) = 0.04 ( 0.99) + 0.96 ( 0.02) = 0.059

Pr (N) = 0.04 (0.01) + 0.96 (0.98) = 0.941

Pr (DP) = 0.04 * 0.99 = 0.0396

Pr (HN) = 0.96 * 0.98 = 0.9408

Probability of a positive test rest result is 0.059.

2. To find : Pr (D|P) = ............(By definition of conditional probability)

=

Given a randomly selected adult is tested and the result is positive, the probability that the individual has the disease is 0.671.

3. To find Pr ( H|N) =

=

Given that a randomly selected adult is tested and the result is negative, the probability that the individual does not have the disease is 0.9996.