Question:
Suppose that there is a chance for a newly constructed house to collapse whether the design is faulty or not. The chance that the design is faulty is 10\%. The chance that the house will collapse if the design is faulty is 95\% and otherwise it is 45\%. It is seen that the house collapsed. What is the probability that the design is faulty?
Suppose that there is a chance for a newly constructed house to collapse whether the design is faulty or not. The chance that the design is faulty is 10\%. The chance that the house will collapse if the design is faulty is 95\% and otherwise it is 45\%. It is seen that the house collapsed. What is the probability that the design is faulty?
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Apply Bayes' Theorem for conditional probability.
Updated On: Mar 24, 2026
- 0.91
- 0.091
- 0.19
- 0.019
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The Correct Option is C
Solution and Explanation
Step 1: Define events and probabilities.
\( P(F) = 0.10, P(F') = 0.90, P(C|F) = 0.95, P(C|F') = 0.45 \) Step 2: Use Bayes' Theorem: \( P(F|C) = \frac{P(C|F) P(F){P(C)} \).}
Step 3: Calculate \( P(C) \) using total probability.
\( P(C) = P(C|F) P(F) + P(C|F') P(F') = (0.95)(0.10) + (0.45)(0.90) = 0.095 + 0.405 = 0.50 \) Step 4: Calculate \( P(F|C) \).
\( P(F|C) = \frac{(0.95)(0.10)}{0.50} = \frac{0.095}{0.50} = 0.19 \)
\( P(F) = 0.10, P(F') = 0.90, P(C|F) = 0.95, P(C|F') = 0.45 \) Step 2: Use Bayes' Theorem: \( P(F|C) = \frac{P(C|F) P(F){P(C)} \).}
Step 3: Calculate \( P(C) \) using total probability.
\( P(C) = P(C|F) P(F) + P(C|F') P(F') = (0.95)(0.10) + (0.45)(0.90) = 0.095 + 0.405 = 0.50 \) Step 4: Calculate \( P(F|C) \).
\( P(F|C) = \frac{(0.95)(0.10)}{0.50} = \frac{0.095}{0.50} = 0.19 \)
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