Problem 16 - Calculus for Biology and Medicine

Question

A screening test for a disease shows a positive result in \(92 \%\) of all cases when the disease is actually present and in \(7 \%\) of all cases when it is not. Assume that the prevalence of the disease is 1 in 600 . If the test is administered to a randomly chosen individual, what is the probability that the result is positive?

Step-by-Step Solution

Answer

The probability of a positive test is approximately 7.14%.

Step 1: Define the Probabilities

Let's establish the given probabilities. Let event \(D\) represent the presence of the disease (disease is present), and \(\overline{D}\) represent the absence of the disease. - Probability the person has the disease, \(P(D) = \frac{1}{600}\).- Probability the person does not have the disease, \(P(\overline{D}) = 1 - P(D) = \frac{599}{600}\).Let \(T\) represent a positive test result. - Probability of a positive test given the disease is present, \(P(T|D) = 0.92\).- Probability of a positive test given the disease is not present, \(P(T|\overline{D}) = 0.07\).

Step 2: Use Law of Total Probability

To find the probability of a positive test result, \(P(T)\), use the law of total probability:\[P(T) = P(T|D)P(D) + P(T|\overline{D})P(\overline{D})\]Substitute the known values:\[P(T) = (0.92) \left(\frac{1}{600}\right) + (0.07) \left(\frac{599}{600}\right)\]

Step 3: Calculate Each Term

Calculate each term: - \(P(T|D)P(D) = 0.92 \times \frac{1}{600} = \frac{0.92}{600}\)- \(P(T|\overline{D})P(\overline{D}) = 0.07 \times \frac{599}{600} = \frac{41.93}{600}\)

Step 4: Solve for Total Probability

Combine the results to get the probability of a positive test:\[P(T) = \frac{0.92}{600} + \frac{41.93}{600} = \frac{0.92 + 41.93}{600} = \frac{42.85}{600}\]Decimal approximation gives:\[P(T) = 0.071417\]

Step 5: Confirm the Final Result

The final probability that a randomly-chosen individual has a positive test result is approximately 0.071417, or 7.14%.

Key Concepts

Bayes' Theorem

Bayes' Theorem is a fundamental concept in probability theory. It provides a way to update the probability estimate for a hypothesis based on new evidence. This theorem is particularly powerful in the context of medical testing, where it helps determine the probability that a patient has a disease given the results of a test.

The formula for Bayes' Theorem is: \[P(D|T) = \frac{P(T|D) \cdot P(D)}{P(T)}\]where:

• \(P(D|T)\) is the probability of the disease given a positive test result (posterior probability).
• \(P(T|D)\) is the probability of a positive test if the disease is present.
• \(P(D)\) is the initial probability of the disease (prior probability).
• \(P(T)\) is the total probability of a positive test result.

Bayes' Theorem shows how prior knowledge of the disease's prevalence (how common the disease is in the general population) influences the diagnosis after testing. It helps in assessing how convincing a test result is in the context of actual disease presence.

Law of Total Probability

The Law of Total Probability is a key principle used to determine the overall probability of an outcome by considering all possible ways that outcome can occur. This law is particularly useful when you are analyzing problems involving conditional probabilities.

In our exercise, it is used to find the total probability of a positive test result. This law states:\[P(T) = P(T|D) \cdot P(D) + P(T|\overline{D}) \cdot P(\overline{D})\]Here, the total probability of a positive test \(P(T)\) is calculated by considering both scenarios: the disease is present \(D\), and the disease is absent \(\overline{D}\).

This approach ensures that all possibilities are accounted for, giving us an accurate picture of how likely a positive test result is. It splits the complexity of the likelihood of a positive test into simpler, manageable parts. By adding together the probabilities from each scenario, we derive the overall likelihood of getting a positive result.

Conditional Probability

Conditional Probability provides a measure of the probability of an event occurring given that another event has already occurred. It is crucial in assessing scenarios where events are interconnected.

In this problem, conditional probabilities are used to assess the likelihood of a positive test result given whether the disease is present or absent. The notation \(P(T|D)\) reads as "the probability of having a positive test \(T\) given the presence of the disease \(D\)". Similarly, \(P(T|\overline{D})\) is "the probability of a positive test given the disease is absent".

These probabilities provide a deeper understanding of the test's reliability in different circumstances. Conditional probabilities allow us to determine how much trust to put in a test result, adjusting for the actual situation at hand. They are vital for interpreting the effectiveness and accuracy of diagnostic tests in any scenario.