Suppose that there is a chance for a newly constructed building to collapse, whether the design is faulty or not. The chance that the design is faulty is 10%. The chance that the building collapses is 95% if the design is faulty, otherwise it is 45%. If it is seen that the building has collapsed, then what is the probability that it is due to faulty design?
0.19
Calculating the Probability of Faulty Design Given Building Collapse
This problem involves calculating a conditional probability using Bayes' Theorem. We are given probabilities related to a building's design being faulty and the building collapsing under different design conditions. We need to find the probability that the design was faulty, given that the building has collapsed.
Defining Events and Given Probabilities
Let's define the events:
- Let \( F \) be the event that the design is faulty.
- Let \( NC \) be the event that the design is not faulty.
- Let \( C \) be the event that the building collapses.
We are given the following probabilities:
- Probability of faulty design: \( P(F) = 10\% = 0.10 \)
- Probability of non-faulty design: \( P(NC) = 1 - P(F) = 1 - 0.10 = 0.90 \)
- Probability of collapse given faulty design: \( P(C|F) = 95\% = 0.95 \)
- Probability of collapse given not faulty design: \( P(C|NC) = 45\% = 0.45 \)
Our goal is to find the probability that the design is faulty, given that the building has collapsed, which is \( P(F|C) \).
Applying Bayes' Theorem
Bayes' Theorem is the appropriate formula to use when we want to find the probability of an event (faulty design) given that another event (collapse) has occurred, and we know the reverse conditional probability (\( P(C|F) \)) and prior probabilities (\( P(F) \)).
Bayes' Theorem states:
\( P(F|C) = \frac{P(C|F) P(F)}{P(C)} \)
To use this formula, we first need to calculate the total probability of the building collapsing, \( P(C) \).
Calculating the Total Probability of Collapse \( P(C) \)
The building can collapse in two mutually exclusive scenarios: either the design was faulty and it collapsed, or the design was not faulty and it collapsed. We can find the total probability of collapse using the Law of Total Probability:
\( P(C) = P(C \cap F) + P(C \cap NC) \)
Using the definition of conditional probability \( P(A \cap B) = P(A|B) P(B) \), we can rewrite this as:
\( P(C) = P(C|F) P(F) + P(C|NC) P(NC) \)
Now, substitute the given probability values into the formula:
\( P(C) = (0.95)(0.10) + (0.45)(0.90) \)
\( P(C) = 0.095 + 0.405 \)
\( P(C) = 0.500 \)
So, the total probability of the building collapsing is 0.50 or 50%.
Calculating the Probability of Faulty Design Given Collapse \( P(F|C) \)
Now that we have \( P(C) \), we can plug all the necessary values into Bayes' Theorem:
\( P(F|C) = \frac{P(C|F) P(F)}{P(C)} \)
\( P(F|C) = \frac{(0.95)(0.10)}{0.500} \)
\( P(F|C) = \frac{0.095}{0.500} \)
To simplify the fraction, we can multiply the numerator and denominator by 1000:
\( P(F|C) = \frac{0.095 \times 1000}{0.500 \times 1000} = \frac{95}{500} \)
Or, using the decimal form directly:
\( P(F|C) = \frac{0.095}{0.500} = 0.19 \)
Thus, the probability that the design was faulty, given that the building has collapsed, is 0.19.
| Event | Probability |
|---|---|
| Design is Faulty (F) | \( P(F) = 0.10 \) |
| Design is Not Faulty (NC) | \( P(NC) = 0.90 \) |
| Collapse given Faulty Design (C|F) | \( P(C|F) = 0.95 \) |
| Collapse given Not Faulty Design (C|NC) | \( P(C|NC) = 0.45 \) |
| Total Probability of Collapse (C) | \( P(C) = 0.50 \) |
| Faulty Design given Collapse (F|C) | \( P(F|C) = 0.19 \) |
Conclusion
Based on the calculations using Bayes' Theorem, if the building has collapsed, the probability that the design was faulty is 0.19.
Revision Table: Building Collapse Probability
| Concept | Formula Used | Application in Problem |
|---|---|---|
| Bayes' Theorem | \( P(A|B) = \frac{P(B|A) P(A)}{P(B)} \) | Used to find \( P(F|C) \) |
| Law of Total Probability | \( P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + ... \) (for mutually exclusive, exhaustive events \( A_i \)) | Used to find \( P(C) \) where events are F and NC. \( P(C) = P(C|F)P(F) + P(C|NC)P(NC) \) |
| Conditional Probability Definition | \( P(A \cap B) = P(A|B)P(B) \) or \( P(A|B) = \frac{P(A \cap B)}{P(B)} \) | Used implicitly in the Law of Total Probability step. |
Additional Information: Understanding Conditional Probability and Bayes' Theorem
Conditional Probability: This is the probability of an event occurring given that another event has already occurred. It is written as \( P(A|B) \), which means the probability of event A happening given that event B has happened.
Bayes' Theorem: This theorem is fundamental in probability theory and statistics. It describes how to update the probability of a hypothesis (like the design being faulty) based on new evidence (like the building collapsing). It's particularly useful when we know the probability of the evidence given the hypothesis (\( P(C|F) \)) and the prior probability of the hypothesis (\( P(F) \)), and we want to find the probability of the hypothesis given the evidence (\( P(F|C) \)).
Law of Total Probability: This law helps calculate the total probability of an event that can occur under several different, mutually exclusive conditions. In this problem, collapse \( (C) \) can occur with a faulty design \( (F) \) or a non-faulty design \( (NC) \). Since \( F \) and \( NC \) cover all possibilities for the design and are mutually exclusive, we sum the probabilities of collapse under each condition, weighted by the probability of each condition.
These concepts are crucial in fields like risk assessment, medical testing, machine learning, and engineering, including structural reliability where assessing the probability of failure given different factors (like design flaws) is vital.
Similar Questions
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Important Questions from Probability
- If one event is unaffected by the outcome of another event, the two events are said to be:
1. Dependent
2. Independent
3. Mutually exclusive
4. Mutually inclusive - Let $M$ be a randomly chosen non-empty subset of $S = \{1, 2, 3, \dots, 2026\}$.
Which of the following is the probability that the product of all the elements of $M$ is even? - A fair coin is tossed 6 times. The probability of getting at least 5 heads is _________________. (rounded off to two decimal places)
- Consider a random experiment of rolling a fair die. The probability of getting an odd number or a number less than 4 is ______ (Rounded off to two decimal places).
- In a null hypothesis significance test, the significance level $\alpha$ was set to 0.01. For which one or more of the following $p$ values would the null hypothesis be rejected?
