A test is 95% accurate — what is probability of actually being positive (Bayes)

Question

A disease affects 1% of the population. A test for the disease is 95% accurate — meaning it correctly identifies 95% of diseased people (sensitivity = 95%) and 95% of healthy people (specificity = 95%). If a randomly chosen person tests positive, what is the probability they actually have the disease?

Solution — Step by Step

Let:

$D$ = event that the person has the disease
$D^c$ = event that the person does NOT have the disease
$T^+$ = event that the test is positive

Given:

$P(D) = 0.01$ (1% prevalence)
$P(D^c) = 0.99$
$P(T^+ | D) = 0.95$ (test positive given disease — sensitivity)
$P(T^+ | D^c) = 0.05$ (test positive given no disease — false positive rate = 1 - specificity)

We want $P(D | T^+)$ — the probability of having the disease given a positive test.

Bayes’ Theorem states:

P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)}

We need to calculate $P(T^+)$ using the law of total probability.

P(T^+) = P(T^+ | D) \cdot P(D) + P(T^+ | D^c) \cdot P(D^c)

P(T^+) = (0.95)(0.01) + (0.05)(0.99)

P(T^+) = 0.0095 + 0.0495 = 0.059

P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+)} = \frac{0.95 \times 0.01}{0.059}

P(D | T^+) = \frac{0.0095}{0.059} \approx 0.161

The probability of actually having the disease given a positive test is approximately 16.1%.

Why This Works

This result shocks most people — a 95% accurate test gives only a 16% chance of being positive? The key is base rate (prevalence). When a disease is rare (only 1% have it), even a small false-positive rate creates many more false positives than true positives.

Consider 10,000 people:

100 have the disease → 95 test positive (true positives), 5 test negative (false negatives)
9,900 don’t have the disease → 495 test positive (false positives), 9,405 test negative

Out of $95 + 495 = 590$ positive tests, only 95 are true positives:

P(D | T^+) = \frac{95}{590} \approx 16.1\%

This is why mass screening for rare diseases with even highly accurate tests generates many false alarms. It’s not a flaw in the test — it’s a mathematical reality of low prevalence.

Alternative Method

The natural frequency method (counting approach) is often clearer for intuition:

Imagine 10,000 people. 100 have disease, 9,900 don’t.

True positives: $100 \times 0.95 = 95$
False positives: $9,900 \times 0.05 = 495$
Total positives: $95 + 495 = 590$
Probability truly positive: $95/590 \approx 16.1\%$

This gives the same answer as Bayes’ theorem but is more intuitive. For exam write-ups, use the Bayes formula approach — for personal understanding, the frequency approach is clearer.

Bayes’ theorem problems appear regularly in JEE Main and Class 12 CBSE probability chapter (typically 4-6 marks). The standard structure is always: identify $P(D)$ , $P(T^+|D)$ , $P(T^+|D^c)$ , compute $P(T^+)$ via total probability, then apply Bayes’. Organize the information into a table or tree diagram before writing equations. This prevents errors in identifying which conditional is which.

Common Mistake

The most frequent error is confusing $P(T^+ | D)$ (sensitivity) with $P(D | T^+)$ (positive predictive value). These look similar but are completely different. The question asks for $P(D | T^+)$ , not $P(T^+ | D)$ . Many students read “the test is 95% accurate” and immediately write the answer as 95% — this ignores the base rate entirely. Bayes’ theorem exists precisely to correctly “flip” conditional probabilities. Also watch for: forgetting to include the false positive term in $P(T^+)$ , which would give $P(D|T^+) = 95/95 = 100\%$ — clearly wrong.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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