Suppose it is known that \(1 \%\) of the population have a certain kind of cancer. It is also known that a test for this kind of cancer is positive in \(99 \%\) of the people who have it but is also positive in \(2 \%\) of the people who do not have it. What is the probability that a person who tests positive has cancer of this type?

Conditional probability is the probability of one event happening given that another event has already occurred. In this exercise, we use the notation \( P(A | B) \) to represent the probability of event \( A \) occurring given that event \( B \) has occurred.
Here are the key probabilities from the exercise:

• \( P(T | C) \) is the probability of testing positive given the person has cancer.
• \( P(T | eg C) \) is the probability of testing positive given the person does not have cancer.

We can interpret \( P(T | C) = 0.99 \) to mean that if a person has cancer, there is a 99% chance the test will be positive. Similarly, \( P(T | eg C) = 0.02 \) means that even if a person does not have cancer, there is still a 2% chance the test will be positive.
This concept helps us understand how different conditions affect the probabilities of outcomes, which is crucial for making informed decisions.

The law of total probability is fundamental in probability theory. It allows us to calculate the probability of an event by considering all possible ways that event can occur. In this exercise, we need to find \( P(T) \), the total probability of getting a positive test result.
According to the law of total probability, we can break down \( P(T) \) as follows:
\[ P(T) = P(T | C) \cdot P(C) + P(T | eg C) \cdot P(eg C) \]
Here, we are summing the possibilities of testing positive by accounting for both conditions: having cancer and not having cancer.

• \( P(T | C) = 0.99 \)
• \( P(C) = 0.01 \)
• \( P(T | eg C) = 0.02 \)
• \( P(eg C) = 0.99 \)

Substituting these values into the formula, we get:
\[ P(T) = (0.99 \cdot 0.01) + (0.02 \cdot 0.99) \]
\[ P(T) = 0.0099 + 0.0198 = 0.0297 \]
This value, \( P(T) = 0.0297 \), represents the total probability of testing positive, considering all cases.

Probability theory is the branch of mathematics that deals with the analysis of random events. It provides tools to quantify uncertainty and make reasoned predictions. Central to probability theory are concepts like independent and dependent events, random variables, and probability distributions.
In this exercise, we apply Bayes' Theorem, a powerful tool in probability theory. Bayes' Theorem relates the conditional and marginal probabilities of random events and helps us update beliefs with new evidence. The formula used is:
\[ P(C | T) = \frac{P(T | C) \cdot P(C)}{P(T)} \]
Here, \( P(C | T) \) is the probability of having cancer given a positive test result. By substituting the known values:

• \( P(T | C) = 0.99 \)
• \( P(C) = 0.01 \)
• \( P(T) = 0.0297 \)

We get:
\[ P(C | T) = \frac{0.99 \cdot 0.01}{0.0297} = \frac{0.0099}{0.0297} \approx 0.333 \]
Therefore, the probability that a person who tests positive actually has cancer is about 0.333, or 33.3%. Probability theory helps provide a structured way to make such estimations and informed decisions.