The chance that a software engineer will debug an error X correctly is 80%. The chance that software with the error X will crash after the correct debugging is 30%. The chance of crash of the software by wrong debugging is 70%. A software with the error X crashed. The probability that its error was debugged correctly is:
Understanding the Probability Problem
This question involves calculating conditional probability, specifically using Bayes' Theorem. We are given probabilities related to software debugging (correct or wrong) and the chance of a software crash based on the debugging outcome. We need to find the probability that the debugging was correct, given that the software crashed.
Defining Events and Probabilities
Let's define the events involved:
- Event A: The software engineer debugs the error X correctly.
- Event A': The software engineer debugs the error X wrongly.
- Event B: The software with error X crashes.
Based on the question, we are given the following probabilities:
- Probability of correct debugging, \(P(A) = 80\% = 0.80\).
- Probability of wrong debugging, \(P(A') = 1 - P(A) = 1 - 0.80 = 0.20\).
- Probability of crash given correct debugging, \(P(B | A) = 30\% = 0.30\).
- Probability of crash given wrong debugging, \(P(B | A') = 70\% = 0.70\).
Applying Bayes' Theorem for Conditional Probability
We are asked to find the probability that the error was debugged correctly, given that the software crashed. This is the conditional probability \(P(A | B)\). Bayes' Theorem is the appropriate formula to use here:
\[P(A | B) = \frac{P(B | A) \times P(A)}{P(B)}\]
To use this formula, we first need to calculate the total probability of the software crashing, \(P(B)\). The software can crash in two mutually exclusive ways:
- The error is debugged correctly AND the software crashes (event \(A \cap B\)).
- The error is debugged wrongly AND the software crashes (event \(A' \cap B\)).
The probability of the first case is \(P(A \cap B) = P(B | A) \times P(A)\).
The probability of the second case is \(P(A' \cap B) = P(B | A') \times P(A')\).
The total probability of crashing, \(P(B)\), is the sum of these probabilities:
\[P(B) = P(A \cap B) + P(A' \cap B) = (P(B | A) \times P(A)) + (P(B | A') \times P(A'))\]
Calculating Probabilities
Let's calculate the necessary probabilities:
- Probability of correct debugging and crash: \(P(A \cap B) = P(B | A) \times P(A) = 0.30 \times 0.80 = 0.24\).
- Probability of wrong debugging and crash: \(P(A' \cap B) = P(B | A') \times P(A') = 0.70 \times 0.20 = 0.14\).
- Total probability of crash: \(P(B) = P(A \cap B) + P(A' \cap B) = 0.24 + 0.14 = 0.38\).
Now we can use Bayes' Theorem to find the probability of correct debugging given the crash, \(P(A | B)\):
\[P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.38}\]
To express this as a fraction, we can multiply the numerator and denominator by 100:
\[\frac{0.24}{0.38} = \frac{24}{38}\]
This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2:
\[\frac{24 \div 2}{38 \div 2} = \frac{12}{19}\]
Comparing with Options
The calculated probability that the error was debugged correctly, given the software crashed, is \(\dfrac{12}{19}\). Let's compare this with the given options:
- Option 1: \(\dfrac{7}{12}\)
- Option 2: \(\dfrac{7}{19}\)
- Option 3: \(\dfrac{6}{19}\)
- Option 4: \(\dfrac{12}{19}\)
Our calculated value matches Option 4.
| Event | Probability |
|---|---|
| Correct Debugging (A) | \(P(A) = 0.80\) |
| Wrong Debugging (A') | \(P(A') = 0.20\) |
| Crash given Correct Debugging (B|A) | \(P(B|A) = 0.30\) |
| Crash given Wrong Debugging (B|A') | \(P(B|A') = 0.70\) |
| Correct Debugging AND Crash (\(A \cap B\)) | \(P(A \cap B) = P(B|A) \times P(A) = 0.30 \times 0.80 = 0.24\) |
| Wrong Debugging AND Crash (\(A' \cap B\)) | \(P(A' \cap B) = P(B|A') \times P(A') = 0.70 \times 0.20 = 0.14\) |
| Total Probability of Crash (B) | \(P(B) = P(A \cap B) + P(A' \cap B) = 0.24 + 0.14 = 0.38\) |
| Probability of Correct Debugging given Crash (A|B) | \(P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{0.24}{0.38} = \dfrac{12}{19}\) |
Revision Table: Probability Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Conditional Probability | The probability of an event occurring given that another event has already occurred. | \(P(A|B) = \dfrac{P(A \cap B)}{P(B)}\) |
| Bayes' Theorem | Relates conditional probabilities. Used to update the probability of a hypothesis based on new evidence. | \(P(A|B) = \dfrac{P(B|A) \times P(A)}{P(B)}\) |
| Law of Total Probability | Used to find the total probability of an event by considering all possible mutually exclusive scenarios leading to that event. | If \(A_1, A_2, \dots, A_n\) are mutually exclusive and exhaustive events, then \(P(B) = \sum_{i=1}^n P(B|A_i) \times P(A_i)\). |
| Intersection of Events | The event where both A and B occur (\(A \cap B\)). For independent events, \(P(A \cap B) = P(A) \times P(B)\). For dependent events, \(P(A \cap B) = P(A) \times P(B|A)\) or \(P(B) \times P(A|B)\). | \(P(A \cap B)\) |
Additional Information on Software Probability
Probability concepts like conditional probability and Bayes' Theorem are useful in various real-world applications, including software development and testing. Understanding the likelihood of certain events, such as a software crash given a specific action like debugging, helps engineers assess risks and improve processes.
- In software testing, probabilities can model the chance of finding a bug given a certain test strategy.
- Reliability engineering uses probability to predict the likelihood of system failures over time.
- Root cause analysis often involves working backward from an event (like a crash) to determine the most probable cause, which is exactly what Bayes' Theorem helps us do.
- This problem highlights how the outcome of an action (debugging correctly or wrongly) influences the subsequent chance of failure (crashing).
Important Questions from Probability of Random Experiments
A six faced die is a biased one. It is thrice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in the two throws is even is
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From two well shuffled pack of cards what is the probability of getting one Jack from the first one and a King from the second?
If A is an event of getting 13 by throwing two unbiased six-faced dice, then A is called
One summer, Peter visits 4 villages (A, B, C and D) in a random order. Find the probability that he visits (i) A before B (ii) A before B and B before C respectively?
