Question

A chemist wishes to detect an impurity in a certain compound that she is making. There is a test that detects an impurity with probability \(0.90\); however, this test indicates that an impurity is there when it is not about \(5 \%\) of the time. The chemist produces compounds with the impurity about \(20 \%\) of the time. A compound is selected at random from the chemist's output. The test indicates that an impurity is present. What is the conditional probability that the compound actually has the impurity?

Short answer

The conditional probability that the compound actually has the impurity, given that the test indicates an impurity is present, is approximately \(0.818\) or \(81.8 \%\).

Step-by-step solution

Identify the Known Probabilities

Firstly, identify the known probabilities from the problem statement: \n\n1. Probability that a compound has an impurity, denoted \( P(I) \), is 0.20. \n2. Probability that the test detects the impurity, given that the impurity is there, denoted \( P(T | I) \), is 0.90. This is a true positive. \n3. Probability that the test detects the impurity, given no impurity, denoted \( P(T | \neg I) \), is 0.05. This is a false positive.

Apply Bayes' Theorem

We apply Bayes’ theorem to find \( P(I | T) \) (probability that a compound has an impurity given that the test is positive). \n\nBayes' theorem can be expressed as: \n\n\[ P(I | T) = \frac{P(T | I) \cdot P(I)}{P(T | I) \cdot P(I) + P(T | \neg I) \cdot P(\neg I)} \] \n\nWe know \( P(\neg I) \), the probability that a compound doesn't have an impurity, is 1 - \( P(I) \) = 0.80, because it is the complement of \( P(I) \).

Substitute the Values

Substitute the values into the equation: \n\n\[ P(I | T) = \frac{0.90 \cdot 0.20}{(0.90 \cdot 0.20) + (0.05 \cdot 0.80)} \]

Compute the Result

Perform the computation to get the result. \n\n\[ P(I | T) = \frac{0.18}{0.18 + 0.04} = \frac{0.18}{0.22} = 0.818 \]