Q.A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Concept: Conditional Probability (Bayes' Theorem)

Let $D$ = has disease, $T^+$ = positive test.

Given: $P(T^+ \mid D) = 0.99$, $P(T^+ \mid \bar{D}) = 0.005$, $P(D) = 0.001$.

Step 1: Find $P(T^+)$ using the law of total probability:

Step 2: Apply Bayes' Theorem:

Step 3: Simplify: …

Why Bayes’ theorem is the natural tool

We are asked: Given a positive test, what is the chance the person truly has the disease? This is a classic inverse probability problem. The test’s accuracy is given “forward” (if diseased → positive with 99% chance), but we need the reverse direction. Bayes’ theorem is designed exactly for this: it lets us “flip” conditional probabilities using the base rate (prevalence) of the disease.

The key insight: even a very good test can be misleading when the condition is rare. Most positive results will come from the large number of healthy people, not from the few who are actually sick.

Step-by-step solution

1. Define the events clearly

Let $D$ = event that the person has the disease.

Let $T^+$ = event that the test result is positive.

We are given:

• Prevalence: $P(D) = 0.001$ (0.1% of population)
• Test sensitivity (true positive rate): $P(T^+ \mid D) = 0.99$
• False positive rate: $P(T^+ \mid \text{not } D) = 0.005$ (0.5% of healthy people test positive)

2. Write Bayes’ theorem for this situation

The denominator $P(T^+)$ is the total probability of a positive test, which comes from two sources: diseased people who test positive, and healthy people who test positive.

3. Compute the total probability of a positive test

We know $P(\text{not } D) = 1 - 0.001 = 0.999$.

So:

4. Apply Bayes’ theorem

Now simplify: …

Method: Bayes' Theorem with a rare base rate (the false-positive trap)

Use this for medical-test and screening problems: the test's accuracy is given forward (diseased → positive), but you are asked the reverse (positive → diseased), and the condition is rare.

Steps

Step 1: Write the prior (prevalence) and its complement.

$P(D)$ is the base rate of the condition; $P(\bar D) = 1 - P(D)$ is the healthy fraction. Because $P(D)$ is tiny, $P(\bar D)$ is close to $1$ — keep both, do not carelessly round it to $1$.

Step 2: Record the two likelihoods.

$P(E \mid D)$ is the true-positive rate (sensitivity); $P(E \mid \bar D)$ is the false-positive rate.

Step 3: Total probability of a positive result. …

Common Mistakes

Mistake 1: Reading "99% effective" as a 99% chance of disease after a positive test.

Why it's wrong: that ignores the base rate — the disease is present in only $0.1\%$ of people. Correct approach: weight by prevalence via Bayes' theorem; the answer is about $0.165$, not $0.99$.

Mistake 2: Leaving out the false-positive contribution to $P(T^+)$.

Why it's wrong: most positives actually come from the huge healthy group. Correct approach: $P(T^+)=(0.99)(0.001)+(0.005)(0.999)=0.005985$. …