Explain bayes theorem of probability with example

Bayes' theorem is a fundamental concept in probability theory that describes how to update the probability of a hypothesis based on new evidence. It's expressed as:

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$

Where:

• $P(A|B)$ is the posterior probability: the probability of hypothesis A given evidence B.
• $P(B|A)$ is the likelihood: the probability of evidence B given hypothesis A.
• $P(A)$ is the prior probability: the initial probability of hypothesis A before observing evidence B.
• $P(B)$ is the probability of the evidence: the overall probability of observing evidence B.

Example:
Suppose a medical test for a rare disease is 99% accurate (meaning it correctly identifies 99% of people who have the disease and correctly identifies 99% of people who don't have the disease). The disease affects 0.1% of the population. If a person tests positive, what is the probability they actually have the disease?

Let A be the event that a person has the disease, and B be the event that the person tests positive.

• $P(A) = 0.001$ (prior probability of having the disease)
• $P(B|A) = 0.99$ (probability of testing positive given you have the disease - true positive rate)
• $P(\text{not } A) = 0.999$ (prior probability of not having the disease)
• $P(B|\text{not } A) = 0.01$ (probability of testing positive given you don't have the disease - false positive rate)

We need to find $P(A|B)$. First, calculate $P(B)$:
$P(B) = P(B|A)P(A) + P(B|\text{not } A)P(\text{not } A)$
$P(B) = (0.99 \times 0.001) + (0.01 \times 0.999)$
$P(B) = 0.00099 + 0.00999 = 0.01098$

Now, apply Bayes' theorem:
$P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{0.99 \times 0.001}{0.01098} = \frac{0.00099}{0.01098} \approx 0.09016$

So, even with a positive test, the probability of actually having the disease is only about 9.02%, due to the rarity of the disease.