An independent research organization is trying to estimate the probability that an accident at a nuclear power plant will result in radiation leakage. The types of accidents possible at the plant are, fire hazards, mechanical failure, or human error. The research organization also knows that two or more types of accidents cannot occur simultaneously. According to the studies carried out by the organization, the probability of a radiation leak in case of a fire is 20%, the probability of a radiation leak in case of a mechanical 50%, and the probability of a radiation leak in case of a human error is 10%. The studies also showed the following; The probability of a radiation leak occurring simultaneously with a fire is 0.1%. The probability of a radiation leak occurring simultaneously with a mechanical failure is 0.15%. The probability of a radiation leak occurring simultaneously with a human error is 0.12%. On the basis of the information available, answer the questions below: 2.1 What are the probabilities of a fire, a mechanical failure, and a human error respectively? 2.2 What is the probability of a radiation leak? 2.3 Suppose there has been a radiation leak in the reactor for which the definite cause is not known. What is the probability that it has been caused by:
The Correct Answer and Explanation is:
To answer the questions, we will define the events and use Bayes’ Theorem and the Law of Total Probability.
Let:
- FFF: Fire hazard
- MMM: Mechanical failure
- HHH: Human error
- RRR: Radiation leak
We’re given:
- P(R∣F)=0.2P(R|F) = 0.2P(R∣F)=0.2, P(R∩F)=0.001P(R \cap F) = 0.001P(R∩F)=0.001
- P(R∣M)=0.5P(R|M) = 0.5P(R∣M)=0.5, P(R∩M)=0.0015P(R \cap M) = 0.0015P(R∩M)=0.0015
- P(R∣H)=0.1P(R|H) = 0.1P(R∣H)=0.1, P(R∩H)=0.0012P(R \cap H) = 0.0012P(R∩H)=0.0012
Also, F,M,HF, M, HF,M,H are mutually exclusive and exhaustive, meaning:P(F)+P(M)+P(H)=1P(F) + P(M) + P(H) = 1P(F)+P(M)+P(H)=1
2.1. Probabilities of fire, mechanical failure, and human error
Using:P(R∩A)=P(R∣A)⋅P(A)⇒P(A)=P(R∩A)P(R∣A)P(R \cap A) = P(R|A) \cdot P(A) \Rightarrow P(A) = \frac{P(R \cap A)}{P(R|A)}P(R∩A)=P(R∣A)⋅P(A)⇒P(A)=P(R∣A)P(R∩A)
So:
- P(F)=0.0010.2=0.005P(F) = \frac{0.001}{0.2} = 0.005P(F)=0.20.001=0.005
- P(M)=0.00150.5=0.003P(M) = \frac{0.0015}{0.5} = 0.003P(M)=0.50.0015=0.003
- P(H)=0.00120.1=0.012P(H) = \frac{0.0012}{0.1} = 0.012P(H)=0.10.0012=0.012
Total: 0.005+0.003+0.012=0.020.005 + 0.003 + 0.012 = 0.020.005+0.003+0.012=0.02.
Since P(F)+P(M)+P(H)=1P(F) + P(M) + P(H) = 1P(F)+P(M)+P(H)=1, we divide each by 0.02 to normalize.
Thus:
- P(F)=0.0050.02=0.25P(F) = \frac{0.005}{0.02} = 0.25P(F)=0.020.005=0.25
- P(M)=0.0030.02=0.15P(M) = \frac{0.003}{0.02} = 0.15P(M)=0.020.003=0.15
- P(H)=0.0120.02=0.60P(H) = \frac{0.012}{0.02} = 0.60P(H)=0.020.012=0.60
2.2. Probability of a radiation leak
Use the Law of Total Probability:P(R)=P(R∣F)P(F)+P(R∣M)P(M)+P(R∣H)P(H)P(R) = P(R|F)P(F) + P(R|M)P(M) + P(R|H)P(H)P(R)=P(R∣F)P(F)+P(R∣M)P(M)+P(R∣H)P(H)P(R)=(0.2)(0.25)+(0.5)(0.15)+(0.1)(0.60)=0.05+0.075+0.06=0.185P(R) = (0.2)(0.25) + (0.5)(0.15) + (0.1)(0.60) = 0.05 + 0.075 + 0.06 = 0.185P(R)=(0.2)(0.25)+(0.5)(0.15)+(0.1)(0.60)=0.05+0.075+0.06=0.185
2.3. Posterior probabilities using Bayes’ Theorem
We want:
- P(F∣R)=P(R∣F)P(F)P(R)=0.2⋅0.250.185≈0.270P(F|R) = \frac{P(R|F)P(F)}{P(R)} = \frac{0.2 \cdot 0.25}{0.185} ≈ 0.270P(F∣R)=P(R)P(R∣F)P(F)=0.1850.2⋅0.25≈0.270
- P(M∣R)=0.5⋅0.150.185≈0.405P(M|R) = \frac{0.5 \cdot 0.15}{0.185} ≈ 0.405P(M∣R)=0.1850.5⋅0.15≈0.405
- P(H∣R)=0.1⋅0.600.185≈0.324P(H|R) = \frac{0.1 \cdot 0.60}{0.185} ≈ 0.324P(H∣R)=0.1850.1⋅0.60≈0.324
Summary & Explanation
This problem applies foundational concepts of probability, particularly Bayes’ Theorem and the Law of Total Probability, to analyze the causes and likelihood of radiation leaks at a nuclear power plant. Given three mutually exclusive types of accidents—fire, mechanical failure, and human error—the problem provides the probabilities of radiation leaks given each type of accident, as well as the joint probabilities of those events occurring together.
First, we calculate the prior probabilities of each type of accident. Using the formula P(A)=P(R∩A)P(R∣A)P(A) = \frac{P(R \cap A)}{P(R|A)}P(A)=P(R∣A)P(R∩A), we find initial unnormalized probabilities for each event. However, these add up to only 0.02, so we normalize them to sum to 1, yielding:
- Fire: 25%
- Mechanical failure: 15%
- Human error: 60%
Next, we calculate the overall probability of a radiation leak, using the Law of Total Probability. This law sums the conditional probabilities of leaks weighted by the probability of each accident type. We find that P(R)=0.185P(R) = 0.185P(R)=0.185, or an 18.5% chance of a radiation leak from any cause.
Lastly, using Bayes’ Theorem, we reverse the condition to find the posterior probability of each accident type given a radiation leak has occurred. Surprisingly, the most likely cause turns out to be a mechanical failure (≈40.5%), followed by human error (≈32.4%) and then fire (≈27.0%). These results highlight how mechanical failures, although less frequent, pose a higher leak risk.
This structured probabilistic approach supports better risk assessment and preventive measures in nuclear plant operations.
