A new test for detecting a disease has been developed. Medical trials show that 92% o patients with the disease show a positive test result. If 4% of the population have the disease, determine the probability (to 3 decimal places) that:
a.) a radnomly selected person shows a positive result to the test.
b.) A person showing a positive result suffers from the disease.
To determine the probability in both scenarios, we will use conditional probability. Let's break down each question:
a) Probability that a randomly selected person shows a positive result to the test:
To solve this, we need to consider two possibilities: a person has the disease and the test result is positive, or a person does not have the disease and the test result is positive.
Let's calculate the probability of a person having the disease:
Given that 4% of the population has the disease, the probability of a randomly selected person having the disease is 0.04 (or 4/100).
Now, let's calculate the probability of a person not having the disease:
Since 4% of the population has the disease, the remaining 96% does not have the disease. Therefore, the probability of a randomly selected person not having the disease is 0.96 (96/100).
Given that 92% of the patients with the disease show a positive test result, the probability of a positive test result for a person with the disease is 0.92 (92/100).
Now, let's calculate the probability of a positive test result:
The probability of a positive result can be obtained by considering both scenarios:
- The probability of having the disease and a positive result: (Probability of having the disease) * (Probability of a positive result for a person with the disease) = 0.04 * 0.92 = 0.0368
- The probability of not having the disease and a positive result: (Probability of not having the disease) * (Probability of a false positive) = 0.96 * (1 - 0.92) = 0.960 * 0.08 = 0.0768
Now, we sum up both probabilities to obtain the final answer:
Probability of a positive test result = 0.0368 + 0.0768 = 0.1136
Therefore, the probability that a randomly selected person shows a positive result to the test is approximately 0.114 (rounded to 3 decimal places).
b) Probability that a person showing a positive result suffers from the disease:
To solve this, we need to use Bayes' theorem, which states:
P(A|B) = (P(B|A) * P(A)) / P(B), where A and B are events.
In this case, event A represents having the disease, and event B represents a positive test result.
Let's calculate the probability that a person has the disease given a positive test result:
We have already calculated:
- P(B|A): Probability of a positive test result given the disease = 0.92
- P(A): Probability of having the disease = 0.04
- P(B): Probability of a positive test result (calculated in part a) = 0.114
Using Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
= (0.92 * 0.04) / 0.114
= 0.0368 / 0.114
≈ 0.323
Therefore, the probability that a person showing a positive result suffers from the disease is approximately 0.323 (rounded to 3 decimal places).