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 64.8%.

Step 1: Define the Problem Variables

Let's denote the following events for clarity: - \( D \): event that an individual has the disease.- \( D^c \): event that an individual does not have the disease.- \( P_1 \): event that the first test is positive.- \( P_2 \): event that the second test is positive.We're given:- \( P(P_1|D) = 0.95 \)- \( P(P_1|D^c) = 0.10 \)- \( P(P_2|D) = 0.95 \) (since the tests are independent)- \( P(P_2|D^c) = 0.10 \)- \( P(D) = \frac{1}{50} \).The goal is to find \( P(D|P_1 \cap P_2) \): the probability the individual has the disease given two positive tests.

Step 2: Apply Bayes' Theorem

Bayes' theorem provides a way to find the probability of event \( D \) given events \( P_1 \) and \( P_2 \). The formula for conditional probability applying Bayes' theorem is given by:\[P(D|P_1 \cap P_2) = \frac{P(P_1 \cap P_2|D) \cdot P(D)}{P(P_1 \cap P_2)}\]We need to calculate both the numerator and the denominator in this equation.

Step 3: Calculate the Numerator

The numerator \( P(P_1 \cap P_2|D) \cdot P(D) \) can be calculated as:\[P(P_1 \cap P_2|D) = P(P_1|D) \cdot P(P_2|D) = 0.95 \times 0.95 = 0.9025\]Thus, the numerator is:\[P(P_1 \cap P_2|D) \cdot P(D) = 0.9025 \times \frac{1}{50} = 0.01805\]

Step 4: Calculate the Denominator

The denominator \( P(P_1 \cap P_2) \) consists of two parts:\[P(P_1 \cap P_2) = P(P_1 \cap P_2|D) \cdot P(D) + P(P_1 \cap P_2|D^c) \cdot P(D^c)\]We already found the first part. Now, for the second part:\[P(P_1 \cap P_2|D^c) = P(P_1|D^c) \cdot P(P_2|D^c) = 0.10 \times 0.10 = 0.01\]Thus, the complete denominator is:\[P(P_1 \cap P_2) = 0.01805 + 0.01 \times \frac{49}{50} = 0.01805 + 0.0098 = 0.02785\]

Step 5: Calculate the Probability

Now that both the numerator and denominator have been calculated:\[P(D|P_1 \cap P_2) = \frac{0.01805}{0.02785} \approx 0.648\]This result indicates there is approximately a 64.8% probability that the individual has the disease given two positive results.

Key Concepts

Understanding Conditional Probability

Conditional probability helps us understand how likely an event is to occur, given that another event has already happened. In our exercise, the events in question are test results and having a disease. When we say we're interested in the probability of having a disease given two positive tests, we are essentially talking about a conditional probability problem.

To express this mathematically, we use the notation \(P(A|B)\). Here, \(P(A|B)\) represents the probability of event \(A\) occurring given that \(B\) has already occurred. In simpler terms, it's the likelihood of \(A\) knowing \(B\) happened.

• The probability will change based on the information provided by \(B\).
• Conditional probability is crucial in medical testing to understand what test results mean in the context of actual disease presence.

This concept is a foundation for Bayes' Theorem, which allows the calculation of conditional probabilities in more complex scenarios.

Independent Events in Probability

In probability, we often talk about events being independent when the occurrence of one doesn't affect the probability of the other. In our exercise, we're told the second test is independent of the first.

• When events \(A\) and \(B\) are independent, the probability of both occurring together is the product of their individual probabilities: \(P(A \cap B) = P(A) \times P(B)\).
• In the exercise, both tests are independent, meaning the result of the first test does not alter the likelihood of the second test being positive.
• This property simplifies the calculations because we can multiply the probabilities of each positive test to find the joint probability.

Recognizing independence helps streamline our calculations and often simplifies complex problems, especially when using Bayes' Theorem.

Disease Screening: Understanding the Basics

Disease screening tests are designed to detect diseases in individuals who do not yet show symptoms. In our scenario, we deal with a test showing positive results in two different situations: when the disease is present and when it is absent. Though the goal of these tests is accurate diagnosis, no test is perfect.

• Sensitivity: This refers to a test’s ability to correctly identify those with the disease. Here, a 95% sensitivity indicates that 95% of diseased individuals will get a positive result.
• Specificity: This measures a test’s ability to correctly identify those without the disease. A 10% rate for false positives means the test incorrectly identifies 10% of healthy individuals as positive.
• A test with high prevalence (like 1 in 50) means it's more likely that a positive result actually indicates disease.

Understanding the reliability and characteristics of a screening test aids in interpreting results, which can be evaluated using Bayes' Theorem to better predict true disease presence given the test outcomes.