A bakery makes three types of bread: sourdough, rye, and whole wheat. The sourdough accounts for \( 50 \% \) of the bakery's bread production, the rye accounts for \( 30 \% \), and the whole wheat accounts for the emaining \( 20 \% \). The known defect rates for each type of bread are as follows. sourdough has a \( 2 \% \) defect rate, rye has a \( 1.5 \% \) defect rate, and whole wheat has a \( 3 \% \) defect rate. If a customer purchases a oaf and it is found to be defective, the probability that it is a rye bread is: A. \( 25.64 \% \)

To find the probability that a defective loaf of bread is rye, we can use Bayes' theorem. We need to calculate the probability of selecting a rye loaf given that it is defective, denoted as \( P(R | D) \). We can use the following information: - The probability of selecting each type of bread: - \( P(S) = 0.50 \) (sourdough) - \( P(R) = 0.30 \) (rye) - \( P(W) = 0.20 \) (whole wheat) - The defect rates for each type of bread: - \( P(D | S) = 0.02 \) (sourdough) - \( P(D | R) = 0.015 \) (rye) - \( P(D | W) = 0.03 \) (whole wheat) We first need to calculate the total probability of a loaf being defective, \( P(D) \): \[ P(D) = P(D | S) \cdot P(S) + P(D | R) \cdot P(R) + P(D | W) \cdot P(W) \] Substituting the values: \[ P(D) = (0.02 \cdot 0.50) + (0.015 \cdot 0.30) + (0.03 \cdot 0.20) \] Calculating each term: \[ P(D) = (0.01) + (0.0045) + (0.006) = 0.0205 \] Now we can use Bayes' theorem to find \( P(R | D) \): \[ P(R | D) = \frac{P(D | R) \cdot P(R)}{P(D)} \] Substituting the values: \[ P(R | D) = \frac{0.015 \cdot 0.30}{0.0205} \] Calculating the numerator: \[ 0.015 \cdot 0.30 = 0.0045 \] Now substituting back into the equation: \[ P(R | D) = \frac{0.0045}{0.0205} \approx 0.2195 \] To express this as a percentage: \[ P(R | D) \approx 21.95\% \] Thus, the probability that a defective loaf of bread is rye is approximately \( 21.95\% \), which does not match the option \( 25.64\% \). Therefore, the answer is not \( A. 25.64\% \).