Chapter 10: Problem 49
Arsenic exposure and cancer In a certain county, \(2 \%\) of the people have cancer. Of those with cancer, \(70 \%\) have been exposed to high levels of arsenic. Of those without cancer, \(10 \%\) have been exposed. What percentage of the people who have been exposed to high levels of arsenic have cancer? (Hint: Use a tree diagram.)
Short Answer
Step by step solution
Identify Given Probabilities
Use Bayes' Theorem
Calculate Overall Probability of Exposure
Compute Specific Probability Using Bayes’ Theorem
Convert to Percentage
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability
In this exercise, several probabilities are given:
- The probability that a person has cancer, denoted as \( P(C) \), is 0.02, meaning 2% of people in the county have cancer.
- Conversely, the probability that a person does not have cancer, denoted as \( P(eg C) \), can be calculated as \( 1 - P(C) = 0.98 \).
- Additionally, there are conditional probabilities that describe the likelihood of being exposed to arsenic depending on the presence of cancer.
Conditional Probability
- The probability of being exposed to arsenic given that a person has cancer, denoted as \( P(E|C) = 0.70 \). This indicates that 70% of those with cancer have been exposed to high levels of arsenic.
- The probability of arsenic exposure given no cancer is present, \( P(E| eg C) = 0.10 \), meaning 10% of those without cancer have been exposed.
Tree Diagram
Each branch of the tree diagram represents a possible outcome, and by following the branches, we can calculate composite probabilities. Here’s how we can use it:
- Start with the initial probabilities (e.g., having cancer and not having cancer).
- Extend branches to arsenic exposure and no exposure for each starting probability.
- Multiply down the branches to find combined probabilities (e.g., \( P(E \cap C) \) for exposure and cancer).
Law of Total Probability
Here's the breakdown:
- The term \( P(E) \) is split into two scenarios: one where individuals have cancer and are exposed (\( P(E|C) \times P(C) \)) and one where they do not have cancer but still are exposed (\( P(E| eg C) \times P(eg C) \)).
- Mathematically, this means combining both probabilities: \[ P(E) = P(E|C) \times P(C) + P(E| eg C) \times P(eg C) \].