1. Define Events:
- Let \( F_1 \) and \( F_2 \) denote the factories chosen.
- Let \( D_1 \) and \( D_2 \) represent the events that \( B_1 \) and \( B_2 \) are defective, respectively.
2. Given Information:
- \( P(D_1 \mid F_1) = 0.5, \, P(D_2 \mid F_1) = 0.5 \).
- \( P(D_1 \mid F_2) = 0.1, \, P(D_2 \mid F_2) = 0.1 \).
- \( P(F_1) = P(F_2) = 0.5 \).
3. Using Total Probability:
- Compute \( P(D_1) \):
\[
P(D_1) = P(D_1 \mid F_1)P(F_1) + P(D_1 \mid F_2)P(F_2).
\]
Substituting values:
\[
P(D_1) = (0.5)(0.5) + (0.1)(0.5) = 0.25 + 0.05 = 0.3.
\]
4. Conditional Probability \( P(F_1 \mid D_1) \):
- By Bayes' theorem:
\[
P(F_1 \mid D_1) = \frac{P(D_1 \mid F_1)P(F_1)}{P(D_1)} = \frac{(0.5)(0.5)}{0.3} = \frac{0.25}{0.3} = \frac{5}{6}.
\]
5. Conditional Probability \( P(F_2 \mid D_1) \):
- Similarly:
\[
P(F_2 \mid D_1) = \frac{P(D_1 \mid F_2)P(F_2)}{P(D_1)} = \frac{(0.1)(0.5)}{0.3} = \frac{0.05}{0.3} = \frac{1}{6}.
\]
6. Conditional Probability \( P(D_2 \mid D_1) \):
- By the law of total probability:
\[
P(D_2 \mid D_1) = P(D_2 \mid F_1, D_1)P(F_1 \mid D_1) + P(D_2 \mid F_2, D_1)P(F_2 \mid D_1).
\]
- Since \( P(D_2 \mid F_1, D_1) = P(D_2 \mid F_1) = 0.5 \) and \( P(D_2 \mid F_2, D_1) = P(D_2 \mid F_2) = 0.1 \), we get:
\[
P(D_2 \mid D_1) = (0.5)\left(\frac{5}{6}\right) + (0.1)\left(\frac{1}{6}\right).
\]
- Simplify:
\[
P(D_2 \mid D_1) = \frac{5}{12} + \frac{1}{60} = 0.42 \text{ to } 0.44.
\]