Let $C$ be the event that a randomly chosen person has lung cancer. Let $S$ be the event of a person being a smoker.
Suppose that 10% of the population has lung cancer and 20% of the population are smokers. Also, suppose that we know that 70% of all people who have lung cancer are smokers.

Then determine the probability of a person having lung cancer given that the person is a smoker.

Let $P(E \mid F)$ be the probability that $E$ occurs given $F$ occurs. This is called a conditional probability of $E$ given $F$.

Suppose that we know $P(E), P(F)$ and $P(E \mid F)$. Then $P(F \mid E)$ can be computed by Bayes’ theorem (alternatively Bayes’ rule):
\[ P(E \mid F) = \frac{P(E) \cdot P(F \mid E)}{P(F)}.\]

Solution

Given information can be formulated as
\[P(C) = 0.1, P(S) = 0.2, \text{ and } P(S \mid C) = 0.7.\]

The required probability is $P(C \mid S)$. Using Bayes’ rule, we can compute it as follows.
\begin{align*}
P(C \mid S) &= \frac{P(C) \cdot P(S \mid C)}{P(S)}\\[6pt]
&= \frac{(0.1)(0.7)}{0.2}\\[6pt]
&= 0.35
\end{align*}

Remark

The data given here is artificial for educational purpose and is not based on a scientific fact.

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