Question:
Two factories \( F_1 \) and \( F_2 \) produce cricket bats that are labelled. Any randomly chosen bat produced by factory \( F_1 \) is defective with probability 0.5 and any randomly chosen bat produced by factory \( F_2 \) is defective with probability 0.1. One of the factories is chosen at random, and two bats are randomly purchased from the chosen factory. Let the labels on these purchased bats be \( B_1 \) and \( B_2 \). If \( B_1 \) is found to be defective, then the conditional probability that \( B_2 \) is also defective is equal to __________ (round off to 2 decimal places).
Two factories \( F_1 \) and \( F_2 \) produce cricket bats that are labelled. Any randomly chosen bat produced by factory \( F_1 \) is defective with probability 0.5 and any randomly chosen bat produced by factory \( F_2 \) is defective with probability 0.1. One of the factories is chosen at random, and two bats are randomly purchased from the chosen factory. Let the labels on these purchased bats be \( B_1 \) and \( B_2 \). If \( B_1 \) is found to be defective, then the conditional probability that \( B_2 \) is also defective is equal to __________ (round off to 2 decimal places).
Updated On: Jan 25, 2025
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Correct Answer: 0.42 - 0.44
Solution and Explanation
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:
\[
P(D_2 \mid D_1) = \frac{5}{12} + \frac{1}{60} = 0.42 \text{ to } 0.44.
\]
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