Q.By examining the chest X-ray, the probability that TB is detected when a person is actually suffering is $0.99$. The probability of a healthy person diagnosed to have TB is $0.001$. In a certain city, $1$ in $1000$ people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?

Concept: Conditional Probability (Bayes' Theorem)

Let $T$ be the event that a person has TB, and $D$ be the event that the test diagnoses TB.

Given:

$P(D|T) = 0.99$,

$P(D|T^c) = 0.001$,

$P(T) = \frac{1}{1000} = 0.001$.

We need $P(T|D)$.

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

$P(D) = P(D|T)P(T) + P(D|T^c)P(T^c)$

$= (0.99)(0.001) + (0.001)(0.999) = 0.00099 + 0.000999 = 0.001989$.

Step 2: Apply Bayes' Theorem: …

Why Bayes' theorem is the natural tool here

We are given a test (chest X-ray) that has two kinds of accuracy: it catches true cases with high sensitivity (99%), but it also occasionally gives false alarms — a healthy person gets wrongly diagnosed 0.1% of the time. The real twist is that TB is rare: only 1 in 1000 people actually has it. So even a very good test will produce many false positives simply because there are so many more healthy people.

The question asks: Given that the test says "TB", what is the chance the person really has it? That is a classic inverse probability problem — we know $P(\text{positive} \mid \text{TB})$ and $P(\text{positive} \mid \text{healthy})$, but we want $P(\text{TB} \mid \text{positive})$. Bayes' theorem is the only correct way to reverse the conditioning.

Let's define the events clearly and apply it step by step.

Step-by-step solution

1. Define the events

Let $T$ = "person has TB" and $H$ = "person is healthy" (so $H = T^c$).

Let $D$ = "person is diagnosed with TB (test positive)".

From the problem:

• $P(T) = \frac{1}{1000} = 0.001$ (prior probability of TB in the city)
• $P(H) = 1 - 0.001 = 0.999$
• $P(D \mid T) = 0.99$ (sensitivity: test catches 99% of true cases)
• $P(D \mid H) = 0.001$ (false positive rate: 0.1% of healthy people test positive)

2. What we need

We want $P(T \mid D)$ — the probability that a person actually has TB given that the test says so.

3. Apply Bayes' theorem

We already have the numerator: $0.99 \times 0.001 = 0.00099$.

The denominator $P(D)$ is the total probability of a positive test, which can happen in two ways: a true positive (TB person tests positive) or a false positive (healthy person tests positive). By the law of total probability:

Substitute the numbers:

Compute each term:

• $0.99 \times 0.001 = 0.00099$
• $0.001 \times 0.999 = 0.000999$

So $P(D) = 0.00099 + 0.000999 = 0.001989$.

4. Compute the desired probability

Do the division:

$P(T \mid D) \approx 0.4975$ …

Method: Bayes' theorem with a rare base rate

Use this for diagnostic-test problems ("test is $99\%$ accurate; given a positive result, what is the true chance of disease?") where the underlying condition is rare.

Steps

Step 1: Identify the three inputs

The prior/prevalence $P(T)$, the sensitivity $P(D\mid T)$, and the false-positive rate $P(D\mid T^{c})$.

Step 2: Total probability of a positive result

Both a true positive and a false positive feed this denominator. …

Common Mistakes

Mistake 1: Base-rate neglect

Why it's wrong: a $99\%$-accurate test does not give a $99\%$ chance of disease when only $1$ in $1000$ is affected — the rarity dominates. Correct approach: run full Bayes with the prevalence.

Mistake 2: Dropping the false-positive term …