ML Interview Q Series: Calculating System Failure Probability: Leveraging Component Independence and Complementary Events.
Browse all the Probability Interview Questions here.
Question: An electronic system has four components labeled as 1, 2, 3, and 4. The system must be used during a given time period. The probability that component i will fail during that period is f_i for i = 1,...,4. The failures of the components are physically independent of each other. A system failure occurs if component 1 fails or if at least two of the other components fail. Specify an appropriate sample space and determine the probability of a system failure.
Short Compact solution
The sample space can be taken as all 4-tuples (δ1, δ2, δ3, δ4), where δi is 0 if component i fails, and 1 if it does not fail. Each 4-tuple (δ1, δ2, δ3, δ4) is assigned probability r1 r2 r3 r4, where r_i = f_i if δi = 0 and r_i = 1−f_i if δi = 1. A direct way to find the probability that the system fails is to compute the complement of that event (i.e., the probability that the system does not fail) and then subtract from 1.
Let A be the event that none of the four components fail. Let B be the event that component 1 does not fail and exactly one among components 2, 3, and 4 fails. Then A and B cover the scenarios where the system does not fail, so the probability that the system does not fail is P(A) + P(B).
P(A) = (1−f1)(1−f2)(1−f3)(1−f4)
P(B) = (1−f1)[ f2(1−f3)(1−f4) + (1−f2)f3(1−f4) + (1−f2)(1−f3)f4 ]
Hence the probability that the system fails is:
In explicit expanded form, the probability of system failure is:
1 − [(1−f1)(1−f2)(1−f3)(1−f4) + (1−f1){f2(1−f3)(1−f4) + (1−f2)f3(1−f4) + (1−f2)(1−f3)f4}].
Comprehensive Explanation
To solve this problem, we consider the fact that each component has an independent probability of failing. The system fails if component 1 fails or if at least two of components 2, 3, and 4 fail. This structure suggests that instead of directly counting all tuples where this system failure condition is met, it is simpler to calculate the probability of the complementary event (the system does not fail) and subtract it from 1.
The sample space is all possible outcomes of the four components across the time period in question. We denote δi = 0 to indicate that component i fails, and δi = 1 to indicate that it does not fail. Since each of the four components independently can be either 0 or 1, there are 2^4 = 16 possible outcomes (4-tuples).
We assign probabilities to each 4-tuple by multiplying the probability contributions from each component, leveraging independence. Specifically, if component i fails in a particular outcome (δi = 0), that happens with probability f_i, and if it does not fail (δi = 1), that happens with probability 1−f_i.
Because the failure criteria is “component 1 fails OR (at least two of the other three fail),” an equivalent re-interpretation of the no-failure event is “component 1 does not fail AND at most one of components 2, 3, and 4 fails.” We handle these possibilities via two mutually disjoint events:
A: No component fails at all (all four are good).
B: Component 1 is good (δ1=1) and exactly one among components 2, 3, and 4 fails.
We compute P(A) as (1−f1)(1−f2)(1−f3)(1−f4). This covers the scenario where none of the components fail.
Then we compute P(B) by noting that exactly one of components 2, 3, or 4 fails and component 1 does not fail. So, we multiply (1−f1) by the sum of the probabilities that exactly one among 2, 3, and 4 fails. That sum is: f2(1−f3)(1−f4) + (1−f2)f3(1−f4) + (1−f2)(1−f3)f4.
Hence, P(B) = (1−f1)[ f2(1−f3)(1−f4) + (1−f2)f3(1−f4) + (1−f2)(1−f3)f4 ].
Because events A and B are disjoint and they cover all scenarios under which the system does not fail, the probability that the system remains operational (does not fail) is P(A) + P(B). Therefore, the probability that the system fails is:
1 − (P(A) + P(B)).
This logic cleanly captures both ways in which the system could fail (component 1 failing or at least two of 2, 3, 4 failing) by focusing on the complementary perspective that often simplifies computations. The explicit expansion of the final expression follows just by carrying out the required algebra.
When implementing or extending such a model in practice (for instance, if you had more components, or different conditions for failure), the same principle applies. One usually identifies the most straightforward route to counting success cases versus failure cases and uses independence to write down product probabilities.
Follow-up question: Why is using the complement often simpler for this type of probability problem?
This kind of probability problem involves multiple conditions for failure and can lead to a complicated expression if tackled head-on. When two or more events need to be combined with “OR,” directly summing the probabilities can force you to account for their intersections correctly, which can get cumbersome. By looking at the complement (where the system does not fail), you only deal with simpler “AND” conditions (e.g., all components good, or only one fails). These simpler events lend themselves to more direct multiplication of probabilities and less complicated counting.
Follow-up question: Could we have derived the same result by enumerating all cases of system failure?
Yes, you could systematically list all 4-tuples where component 1 fails or at least two of 2, 3, and 4 fail. This direct enumeration would involve identifying all tuples where δ1=0 (component 1 fails) or counting the tuples where exactly two, or exactly three, or all four of 2, 3, and 4 fail. However, merging and combining these cases typically leads to more intricate algebra and a bigger chance for arithmetic errors. Thus, the complement approach is cleaner and less error-prone.
Follow-up question: How would we implement this in a Python code snippet for arbitrary probabilities of failure?
Below is a Python snippet demonstrating how you might compute the system failure probability directly, given any f1, f2, f3, f4:
def system_failure_probability(f1, f2, f3, f4):
import itertools
# We represent each outcome as a 4-tuple of {0,1}, where 0 indicates failure, 1 indicates no failure.
outcomes = list(itertools.product([0,1], repeat=4))
total_prob = 0.0
for outcome in outcomes:
# Probability of this specific outcome:
# If outcome[i] == 0 (i.e., fails), multiply by f_i
# If outcome[i] == 1 (i.e., no fail), multiply by (1 - f_i)
p_outcome = 1.0
for i, delta_i in enumerate(outcome, start=1):
if delta_i == 0:
p_outcome *= (f1 if i == 1 else f2 if i == 2 else f3 if i == 3 else f4)
else:
p_outcome *= ((1 - f1) if i == 1 else (1 - f2) if i == 2 else (1 - f3) if i == 3 else (1 - f4))
# Check if this outcome leads to a system failure:
# Condition: either component 1 fails, or at least two of 2, 3, 4 fail.
# component 1 fails if outcome[0] == 0
# number of fails among 2,3,4 is sum of outcome[1], outcome[2], outcome[3] with 0=fail, 1=no fail
# so the number of fails among 2,3,4 is 3 - (outcome[1] + outcome[2] + outcome[3])
if outcome[0] == 0 or (3 - (outcome[1] + outcome[2] + outcome[3])) >= 2:
total_prob += p_outcome
return total_prob
# Example usage:
p_fail = system_failure_probability(0.2, 0.3, 0.4, 0.1)
print("Probability of system failure:", p_fail)
This code demonstrates the brute force enumeration. Each outcome is explicitly assigned its probability, and then we check if it satisfies the system failure condition. The result is summed over all outcomes that produce a failure. Although this method is conceptually straightforward, it is not as elegant or computationally efficient as the complementary approach, especially as the number of components grows.
Follow-up question: What are some edge cases or boundary values to consider?
If f1 = 0 (component 1 never fails) but the other three have non-zero probabilities, the system still fails only if at least two of 2, 3, and 4 fail. Hence, the probability of system failure would be the probability that at least two out of those three components fail. Another interesting boundary case is if one of the fi = 1, meaning that component i always fails. For example, if f1 = 1, then the system fails with probability 1, because failure of component 1 alone brings down the system. Such edge cases are easily covered by the formula, but they provide good checks to confirm our logic is correct.
Below are additional follow-up questions
How would the analysis change if the component failures were not perfectly independent?
In many real-world systems, component failures may be correlated. For instance, overheating in one component can increase the likelihood of failure in adjacent components. When independence does not hold, the probability of any specific tuple cannot simply be computed as the product of individual failure probabilities. Instead, we must define a joint probability distribution for the four components. One way is to specify a correlation structure (e.g., via copulas or a conditional probability table) and compute probabilities for each tuple accordingly.
A key pitfall is that if you assume independence when there is actually positive correlation, you will underestimate the probability of multiple components failing simultaneously. This can lead to overly optimistic estimates of overall system reliability. In practice, partial correlations often arise from shared operating conditions such as temperature, voltage fluctuations, or mechanical stress. A thorough risk assessment should account for these common factors.
What if the failure probabilities f1, f2, f3, f4 change over time rather than remain constant?
If the probabilities change over time, one must treat the problem dynamically. Instead of a single time window with a fixed failure probability per component, you might break the operational period into smaller intervals, each with its own failure rates. Over each interval, you can compute partial probabilities of failing or surviving that slice of time and then multiply (or integrate) over the intervals. Alternatively, one might model each component using a time-dependent hazard function h(t), which then gives the survival or failure function more explicitly.
A subtle challenge arises in identifying how quickly or slowly the probabilities vary with time. If the time period is short compared to the rate of change in conditions, you might approximate the probabilities as constant. But if the environment changes substantially (e.g., temperature swings, battery depletion, or wear-and-tear effects), ignoring time variation can be a major oversight.
How can we estimate or validate the given failure probabilities f_i in practice?
Estimating f_i often involves analyzing historical data or performing controlled stress tests of each component. One might collect statistics on the frequency of component failures under similar operational conditions and then infer each probability f_i. If test data is limited, Bayesian methods can be used to combine prior knowledge with limited observations.
A real-world pitfall is that laboratory conditions do not always match actual operating conditions (e.g., in stress tests, the environment may be artificially harsher or more benign). Therefore, any mismatch can degrade the reliability of the estimated probabilities. It is crucial to monitor the system in the field and adjust the estimated probabilities if operating conditions turn out to be different from initial assumptions.
What if we only observe the overall system failure events in a black-box manner but not which specific components fail?
Without knowledge of which specific component failed in each event, it is more challenging to isolate each f_i. You might attempt to use system-level data—observing how often the system fails—and combine it with partial testing of individual components or assumptions about how often single failures occur. Some approaches use maximum likelihood estimation under the assumption that failures follow certain distributions or constraints.
An additional subtlety is that if the system fails (e.g., because component 1 fails) you do not necessarily know whether or not other components are also failing at that time. There could be hidden failure states. In many practical settings, detailed logging or diagnostic checks are introduced so that when the system fails, the cause can be at least partially traced. Otherwise, you face identifiability issues: multiple sets of f_i values can produce nearly identical system-level failure patterns, making it hard to pinpoint the exact component-level probabilities.
How does system repair or replacement affect these probabilities?
If the system or individual components are repairable or replaceable within the operational period, the probability model changes because a failed component might be restored, returning the system to a semi-operational or fully operational state. One way to handle this is to employ a discrete-state Markov or semi-Markov process, where states reflect the current health of each component. Transitions occur at random times depending on both failure rates and repair rates.
A pitfall here is that repairs might not restore a component to an “as good as new” condition, but instead an “as bad as old” or somewhere in between. If each repair only partially resets the wear on a component, then the effective failure probability after repair can differ from the initial f_i. Another subtlety is the time it takes to detect and repair the failure; the system could remain in a degraded state for a non-trivial duration, significantly affecting the overall reliability calculations.
How do manufacturing tolerances and environmental conditions factor into these probabilities?
Manufacturing tolerances introduce variability in the nominal performance of components. Two seemingly identical components from the same production batch can exhibit different real-world failure rates, especially under stress. Meanwhile, environmental factors—such as extreme temperatures, humidity, shock, vibration, or voltage spikes—can dramatically change individual f_i values over time.
In practical terms, you might handle this by modeling f_i as a distribution rather than a single point estimate, and then performing a sensitivity analysis or a Monte Carlo simulation. You would draw random samples for each f_i from an appropriate distribution (e.g., Beta distribution if f_i is a probability) and then compute the resulting probability of system failure. By examining how the output varies, engineers gain insight into which components are most sensitive to variance in operating environments.
Could we extend this problem to address partial functionality or performance degradation rather than a strict fail/no-fail model?
Yes, many real systems exhibit partial functionality. For instance, a memory module might still operate but with reduced speed or increased error rates. In such cases, one can move from a binary fail/no-fail model to a multi-state reliability model. Each component can be in one of several states (e.g., fully functional, partially degraded, completely failed), and the system failure criteria might be triggered if enough components are in a degraded or failed state to compromise overall operation.
A frequent pitfall in partial-functionality models is oversimplifying the transition probabilities between states. For instance, a component that enters a degraded state might have a significantly higher chance of outright failure in the near future. Without accurately modeling these state transitions, reliability estimates can be off by a large margin.
What if the definition of "system failure" is more complex, involving performance thresholds instead of strict binary failure conditions?
In some systems, failure is not a binary event but is defined by a performance threshold. For example, a communication system might be considered “failed” if throughput falls below a certain rate. If multiple components share responsibilities, it may be necessary to assess the combined performance. Each component has a probability distribution of performance rather than a single fail probability.
A subtle real-world complication is that interactions between components may be nonlinear. For example, if two out of three channels degrade, the total throughput might not degrade as a simple sum but might trigger a more severe slowdown. Thorough performance modeling often requires a system-level simulation or a carefully constructed analytic model that captures these interdependencies.
Why might we need to consider “mission time” and sequences of failures, rather than just a static single-period probability?
In many practical applications, the system must remain operational for a given “mission time” (e.g., the duration of a spaceflight or a certain shift in a factory). Failures can occur at any point within this time, potentially impacting overall mission success differently depending on the sequence of failures. For instance, if component 1 fails early, the entire mission might be lost immediately, while failure of a secondary component that only gets used toward the end could have delayed or no immediate impact.
Modeling the sequence of events might require more detailed survival analysis or reliability block diagrams that factor in not just whether a component fails within the mission, but also when. If replacements or repairs happen mid-mission, you also need to capture that time aspect. A pitfall is to treat these events as if they occur instantaneously at the start or the end of a mission, which can hide the true risk profile if partial usage of the system is critical.