Question

Only 1 in 1000 adults is aftticted with a particular rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will occur \(99 \%\) of the time, and an individual without the disease will show a positive test result only \(2 \%\) of the time. If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease? Draw a tree diagram for the problem.

Short answer

The probability is approximately 4.72%.

Step-by-step solution

Understand the Problem

We need to find the probability that a person has the disease given a positive test result. This requires Bayes' theorem as it involves conditional probability. The important figures are: probability of having the disease is \( P(D) = \frac{1}{1000} \), probability of testing positive given disease \( P(T^+|D) = 0.99 \), and probability of testing positive without the disease \( P(T^+|eg D) = 0.02 \).

Define Bayes' Theorem

Bayes' Theorem is given by: \[ P(D|T^+) = \frac{P(T^+|D) P(D)}{P(T^+)} \] where \( P(T^+) \) is the total probability of testing positive.

Calculate Total Probability of Testing Positive \(P(T^+)\)

The total probability of testing positive is the sum of testing positive if the person has the disease and if they do not have the disease. This is calculated as: \[ P(T^+) = P(T^+|D)P(D) + P(T^+|eg D)P(eg D) \] Given that \( P(eg D) = 1 - P(D) = \frac{999}{1000} \), we have: \[ P(T^+) = 0.99 \times \frac{1}{1000} + 0.02 \times \frac{999}{1000} \]

Solve for \(P(T^+)\)

Perform the calculation: \[ P(T^+) = 0.99 \times \frac{1}{1000} + 0.02 \times \frac{999}{1000} = 0.00099 + 0.01998 = 0.02097 \]

Calculate \(P(D|T^+)\)

Substitute the values into the Bayes' theorem formula to find \( P(D|T^+) \): \[ P(D|T^+) = \frac{0.99 \times \frac{1}{1000}}{0.02097} = \frac{0.00099}{0.02097} \approx 0.0472 \] This result means the probability that a person has the disease given a positive result is approximately 4.72%.

Key concepts

Conditional Probability

Conditional probability is a fundamental concept in probability theory that asks the question: "Given that one event has occurred, what is the probability that another event will happen?" In this exercise, we're dealing with the probability that an individual has a disease when the test result is positive.

Using the formula for conditional probability, we compute this by looking at the likelihood of a positive test when the disease is present, weighted by how likely it is for the disease to be present in the first place. This is a case where Bayes’ Theorem becomes essential.

• If Event A is having the disease, and Event B is testing positive, conditional probability helps us assess probabilities like \( P(A|B) \), the probability of A given B.
• Bayes’ Theorem, \( P(A|B) = \frac{P(B|A) P(A)}{P(B)} \), provides a way to reverse our conditional probability when direct calculation is complex.
• This theorem shows how to update prior probabilities, like having a disease, with new evidence, like a test result.

Probability Theory

Probability theory is the branch of mathematics that investigates the likelihood of events happening. It’s an essential tool for making sense of random phenomena by assigning a probability value between 0 and 1 to each potential outcome of an event.

In the context of this problem, we're looking at multiple probabilities:

• Probability of having the disease \( P(D) = \frac{1}{1000} \).
• Probability of a positive test given the disease, \( P(T^+|D) = 0.99 \).
• Probability of a positive test without the disease, \( P(T^+|eg D) = 0.02 \).

To find \( P(D|T^+) \), the probability of having the disease given a positive test result, we also compute \( P(T^+) \), the overall probability of testing positive, by considering both accurate positives (correct disease detection) and false positives (erroneous disease detection):

• Correct positives: \( 0.99 \times \frac{1}{1000} \).
• False positives: \( 0.02 \times \frac{999}{1000} \).

These foundational calculations reflect how probability theory is used to interpret data and infer conclusions in uncertain scenarios.

Tree Diagram

A tree diagram is a visual representation used to illustrate all possible outcomes of an event or series of events. It is especially useful in probability to visually map out all potential paths and their associated probabilities.

For this problem, a tree diagram helps display how a person might test positive or negative and whether they have the disease or not.

• Start from the initial branch: Probability of having the disease (\(\frac{1}{1000} \)) and not having the disease (\(\frac{999}{1000}\)).
• Each branch further splits into two: True positive (\(0.99 \)) or false negative (\(0.01 \)) if the disease is present, and false positive (\(0.02 \)) or true negative (\(0.98 \)) if not.
• The diagram captures the entirety of the problem’s event space, making the calculation of total probabilities manageable and visual.

Using a tree diagram helps in systematically navigating through complex probability calculations. When done correctly, it reduces errors and allows for a clear presentation of how every outcome and their probabilities integrate into the overall scenario.