Q.Suppose that the reliability of a HIV test is specified as follows: Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are judged HIV$-$ive but 1% are diagnosed as showing HIV$+$ive. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV$+$ive. What is the probability that the person actually has HIV?

The key idea is Conditional Probability — specifically, Bayes’ theorem, which reverses the conditional probability using the base rate of the disease.

Let $H$ = event that the person has HIV, and $+$ = event that the test is positive.

We are given:

• $P(H) = 0.001$ (0.1% prevalence)
• $P(+ \mid H) = 0.90$ (sensitivity)
• $P(+ \mid H^c) = 0.01$ (false positive rate, since 1% of healthy people test positive)

We want $P(H \mid +)$.

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

Step 2: Apply Bayes’ theorem: …

Why Bayes' theorem is the natural tool

We want $P(\text{HIV} \mid \text{positive})$. That's a conditional probability where the condition (the test result) is observed, but we need to reverse the direction of the given conditional probabilities. The test tells us $P(\text{positive} \mid \text{HIV})$ and $P(\text{negative} \mid \text{no HIV})$, but we need the inverse.

Bayes' theorem is exactly the formula for this reversal. It combines:

• The prior probability of having HIV (the population prevalence: 0.1%)
• The likelihood of a positive test given HIV (90%)
• The total probability of a positive test (which includes both true positives and false positives)

The result is the posterior probability — our updated belief after seeing the positive test.

Step-by-step solution

1. Define the events clearly

Let $H$ = person has HIV, and $T^+$ = test reports positive.

From the problem:

• $P(H) = 0.1\% = 0.001$ (prevalence)
• $P(T^+ \mid H) = 90\% = 0.9$ (sensitivity — true positive rate)
• $P(T^- \mid \overline{H}) = 99\% = 0.99$ (specificity — true negative rate)

Therefore:

• $P(\overline{H}) = 1 - 0.001 = 0.999$
• $P(T^+ \mid \overline{H}) = 1 - 0.99 = 0.01$ (false positive rate)

2. Find the total probability of a positive test

A positive test can happen in two ways:

• The person has HIV and the test correctly detects it: $P(H) \times P(T^+ \mid H)$
• The person does not have HIV and the test falsely says positive: $P(\overline{H}) \times P(T^+ \mid \overline{H})$

By the law of total probability:

Substitute the numbers:

Compute each term:

• True positives: $0.001 \times 0.9 = 0.0009$
• False positives: $0.999 \times 0.01 = 0.00999$

So:

3. Apply Bayes' theorem

We want $P(H \mid T^+)$:

$P(H \mid T^+) = \frac{P(H) \cdot P(T^+ \mid H)}{P(T^+)}$ …

Method: Bayes' Theorem (finding the cause from the observed outcome)

Use this for diagnostic-test problems: you observe a test result and want the probability that the person truly has the condition. The test gives you $P(\text{positive}\mid\text{disease})$ and $P(\text{negative}\mid\text{healthy})$, but you need the reverse, $P(\text{disease}\mid\text{positive})$.

Steps

Step 1: Set up the partition of possible causes.

List the mutually exclusive, exhaustive hypotheses (here: the person has the disease, or is free of it) and write their prior probabilities $P(E_i)$ (these must sum to $1$).

Step 2: Write each likelihood.

For each cause, state $P(A\mid E_i)$ — the probability of the observed outcome (here: the test reports positive) under that cause.

Step 3: Get the total probability of the outcome (the denominator).

By the law of total probability,

Step 4: Apply Bayes' theorem for the cause you want.

$P(E_k\mid A)=\frac{P(E_k)\,P(A\mid E_k)}{\sum_i P(E_i)\,P(A\mid E_i)}.$ …

Common Mistakes

Mistake 1: Thinking a positive result on a "90% / 99%" test means about a 90% chance of disease.

Why it's wrong: this ignores the base rate — only $0.1\%$ of people have HIV, so the false positives from the $99.9\%$ healthy population swamp the true positives. Correct approach: use Bayes' theorem; the answer is only about $8.3\%$.

Mistake 2: Omitting the false-positive term in the denominator.

Why it's wrong: a positive test comes from true positives and false positives. Correct approach: $P(+)=P(H)P(+\mid H)+P(H')P(+\mid H')$. …