Problem 38 - Calculus for Biology and Medicine

Question

A screening test for a disease shows a positive result in \(95 \%\) of all cases when the disease is actually present and in \(10 \%\) of all cases when it is not. If a result is positive, the test is repeated. Assume that the second test is independent of the first test. If the prevalence of the disease is 1 in 50 and an individual tests positive twice, what is the probability that the individual actually has the disease?

Step-by-Step Solution

Answer

The probability that the individual actually has the disease is approximately 66.51%.

Step 1: Define the Events

Let A be the event that the individual has the disease, and B be the event that the test result is positive. We know the following probabilities: \( P(A) = \frac{1}{50} \), \( P(B|A) = 0.95 \), and \( P(B|A') = 0.10 \), where \( A' \) is the event that the individual does not have the disease.

Step 2: Calculate Probability of Positive Tests Given Disease

Since the tests are independent, the probability of two positive results given the disease is present (event A) is \( (P(B|A))^2 = 0.95 \times 0.95 = 0.9025 \).

Step 3: Calculate Probability of Positive Tests Given No Disease

Similarly, the probability of two positive results given the disease is not present (event A') is \( (P(B|A'))^2 = 0.10 \times 0.10 = 0.01 \).

Step 4: Use Bayes' Theorem to Find Probability of Having Disease

We use Bayes' theorem to find the probability that the individual has the disease given they tested positive twice. First, compute \( P(B''|A') \) and \( P(B''|A) \), then use them to find \[ P(A|B'') = \frac{P(B''|A) \cdot P(A)}{P(B'')} \].Where \( P(B'') = P(B''|A) \cdot P(A) + P(B''|A') \cdot P(A') \). Substituting, we find that \( P(A|B'') = \frac{0.9025 \times 0.02}{0.9025 \times 0.02 + 0.01 \times 0.98} \).

Step 5: Perform Calculations and Simplify

Calculate the probabilities: \( P(A) = 0.02 \) and \( P(A') = 0.98 \). Substitute these back into the Bayes' formula: \( P(A|B'') = \frac{0.9025 \times 0.02}{0.9025 \times 0.02 + 0.01 \times 0.98} \approx \frac{0.01805}{0.027125} \approx 0.6651 \).

Key Concepts

Independent Events in Probability

In probability theory, independent events are those whose outcomes do not affect one another. This means that the occurrence of one event has no influence on the occurrence of the other. For example, if you flip a coin, the result of one flip does not impact the result of the next flip, hence each flip is independent. When it comes to testing for diseases, tests are sometimes assumed to be independent, especially if repeated under similar conditions. In our exercise, the result of the first disease test does not affect the result of the second test, making the events independent. This assumption simplifies the calculation of probabilities, because we can find the probability of both tests being positive by multiplying the probability of one positive outcome with itself: \( P(B|A) \) and \( P(B|A') \). Understanding the concept of independent events is crucial, particularly when dealing with conditional probabilities later.

Conditional Probability: Clarifying Uncertainty

Conditional probability is the likelihood of an event occurring given that another event has already occurred. It helps us make sense of the probability of an event in context. The symbol \( P(B|A) \) indicates the probability of event B happening given that A is true. For example, the probability of a test showing positive given that a person actually has the disease is 95%, denoted as \( P(B|A) = 0.95 \). This concept allows us to understand how likely it is to get specific results based on other known information. In our exercise, conditional probability plays a key role in Bayes' Theorem, helping us update our beliefs about an individual's disease status after observing test results. Intuitively, conditional probability acts like a filter that lets us focus on a subset of possibilities given certain information, enhancing our predictions in uncertain situations.

Disease Prevalence: Understanding the Base Rate

Disease prevalence refers to how common the disease is within a particular population. It is usually expressed as a fraction or percentage. Understanding prevalence is essential in evaluating how likely it is for an average person to have the disease without any additional information. In our exercise, the prevalence is given as 1 in 50, which translates to a probability of \( \frac{1}{50} \) or 0.02. This represents the base rate or prior probability of the disease before any tests are taken into account. The prevalence sets the stage for applying Bayes' Theorem. It helps determine the prior probability, which combines with additional information (like test results) to update and refine our estimation. Knowing the prevalence is important because even highly accurate tests can yield misleading results if the actual occurrence of the disease is low. This is why understanding the base rate is critical when interpreting medical test results and making informed decisions.