ML Interview Q Series: Testing Row/Column Event Independence in a Matrix Using Probability Rules
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Fifty different numbers are arranged in a matrix with 5 rows and 10 columns. You pick at random one number from the matrix. Let A be the event that the number comes from an odd-numbered row and B be the event that the number comes from the first five columns. Are the events A and B independent?
Short Compact solution
We pick a number randomly from the matrix of 50 distinct entries. Then:
Total possibilities = 50.
The event A (picking from an odd row) has 3 odd-numbered rows out of 5 total rows, hence 3*10 = 30 possible favorable choices. Therefore P(A) = 30/50.
The event B (picking from one of the first five columns) has 5 columns out of 10 total columns, hence 5*5 = 25 possible favorable choices (because there are 5 rows in total). Therefore P(B) = 25/50.
For the intersection A and B (odd row and one of the first five columns), there are 3 odd rows and 5 columns that match, giving 3*5 = 15 favorable entries. Thus P(A∩B) = 15/50.
Since 15/50 = (30/50)*(25/50), we have P(A∩B) = P(A)P(B). Therefore A and B are independent.
Another way to see this is that picking a row first (uniformly among the 5 rows) and then picking a column (uniformly among the 10 columns) separates the choices in such a way that the choice of row (odd or even) is unaffected by the choice of column, confirming independence.
Comprehensive Explanation
Understanding the setup
We have a 5×10 matrix, making a total of 50 distinct numbers. Each of these 50 numbers is equally likely to be chosen. We define two events:
Event A: “The chosen number lies in one of the odd-numbered rows.” Since the rows are numbered 1,2,3,4,5, the odd rows are 1,3,5, giving 3 odd rows.
Event B: “The chosen number lies in one of the first five columns.” Since the columns are numbered 1 to 10, the first five columns are columns 1,2,3,4,5.
Calculating probabilities
Total number of possible choices = 50.
Probability of A (odd row):
We have 3 odd rows, each with 10 elements. Hence there are 3×10 = 30 favorable outcomes for event A. So P(A) = 30/50.
Probability of B (column in the first five):
We have 5 columns among the first five. Each column has 5 rows (since there are 5 rows in total). Thus there are 5×5 = 25 favorable outcomes for event B. So P(B) = 25/50.
Probability of A∩B (odd row and within the first five columns):
For intersection, we need to be in one of the 3 odd rows and also in one of the 5 columns. Therefore, the number of favorable outcomes is 3×5 = 15. Thus P(A∩B) = 15/50.
The key independence formula
Independence of events A and B in probability requires:
where P(A∩B) is the probability of both A and B occurring simultaneously, while P(A) and P(B) are the individual probabilities of A and B respectively.
Verifying independence
By direct calculation:
Left-hand side: P(A∩B) = 15/50.
Right-hand side: P(A)P(B) = (30/50) × (25/50) = (30×25)/(50×50) = 750/2500 = 15/50.
Because these two quantities are equal, the condition for independence is satisfied, which confirms that A and B are indeed independent.
Alternative viewpoint
One intuitive way to see independence here is to note that the choice of an odd row is essentially a decision about which row block we land in (rows 1,3,5 vs. rows 2,4), while the choice of column (especially among the first five) is a separate choice. Since choosing a row does not affect the columns you have available—each row still spans all 10 columns—knowing whether you’re in an odd row or an even row does not change the relative proportion of columns in the first five columns vs. the last five. This makes the events independent.
Follow-up question 1: What if the numbers in the matrix were not all equally likely to be chosen?
If some numbers had higher probability of being chosen than others, we would not simply count how many entries lie in each row or column. Instead, each entry i would have some probability p_i associated with it. The probability of event A (odd row) would then be the sum of p_i over all entries i in odd rows. Similarly, event B (first five columns) would be the sum of p_i over all entries i in the first five columns. For A and B to be independent, we would need the sum of p_i over entries that are in both odd rows and the first five columns to be the product of the sums for each separate condition. That might no longer hold unless those weights are carefully arranged. In most typical non-uniform weighting scenarios, these events would no longer remain independent.
Follow-up question 2: How does this generalize to different sized matrices or different subsets of rows and columns?
If we had an m×n matrix, and we defined event A as “row belongs to some subset of the m rows” and event B as “column belongs to some subset of the n columns,” then as long as each cell of the matrix is chosen uniformly, the probability of A is the number of rows in that subset times n, divided by m×n. Likewise, the probability of B is the number of columns in that subset times m, divided by m×n. For them to be independent, the intersection must match the product of probabilities. That essentially reduces to: “The chosen row being in the subset is independent of the chosen column being in the subset,” which holds if each cell is chosen uniformly at random. So whenever one picks a row first with 1/m chance per row and then picks a column next with 1/n chance per column, the independence property will hold for any such row-based event and column-based event.
Follow-up question 3: How would you compute expected values under these events?
For instance, suppose each cell of the matrix has a numeric value x_{r,c}, and you wanted to compute the expectation of some function f(A,B). If A and B are events that are independent in a uniform selection scheme, you can factor certain probability terms. However, computing expected values typically involves summing x_{r,c} times its probability. In uniform selection, each x_{r,c} has probability 1/(m×n). If you only restrict to event A (odd row), your expected value might involve summing x_{r,c} over all odd-row cells. The independence concept can also simplify or confirm that knowledge about row selection does not influence how you consider columns, but to do the actual expectation, you typically sum over the entire matrix, weighting each cell’s value by the relevant probability.
Follow-up question 4: Can we have partial independence or conditional independence in such a scenario?
In general probability theory, you can have conditional independence of events given a third event. For example, one might define an event C such as “the chosen number is prime” or “greater than a certain threshold” and ask whether A and B are conditionally independent given C. That would require checking whether P(A∩B | C) = P(A|C) P(B|C). In many practical data scenarios with structured data (like a matrix containing different types of values), you might discover that A and B are independent overall but not necessarily conditionally independent given C, or vice versa. This nuance is relevant to real-world machine learning, where independence assumptions can be broken once you incorporate more context.
All these considerations highlight why independence is such a crucial notion in probability and machine learning: it simplifies our reasoning about complex events, but we must check whether it really holds in each scenario.
Below are additional follow-up questions
Follow-up question 1: What if the matrix has repeated values and we only care about the value drawn, rather than the cell location?
In a scenario where some cells contain the same numerical value, the event “odd row” or “first five columns” still depends on the location in the matrix. However, if we only care about which numerical value is drawn—regardless of where it came from—we might lose information about row or column. This can lead to pitfalls:
Potential Overcounting: If a value v appears multiple times in different rows or columns, simply observing “v was drawn” does not reveal the row or column information. So, calculating P(A) or P(B) becomes trickier if we only have the outcome “the value is v.”
Dependence via Repetition: Two different values could occur a different number of times in odd vs. even rows or in the first five vs. last five columns, creating correlation or dependence that wasn’t present when each cell was unique. If many occurrences of a certain value are in odd rows while other repeated values cluster in even rows, the independence assumptions about row or column might break when we only consider value-based events.
How to handle it: Carefully track which cells contain which values. If you only know the value, you need a conditional probability approach where P(A) or P(B) is replaced by sums over the cells that contain that value, accounting for the frequency distribution of values in each part of the matrix.
Follow-up question 2: How does sampling without replacement affect independence?
Often, one picks numbers without replacement, meaning once a cell’s number is drawn, it’s removed from the pool.
Impact on A and B: If the first drawn number is from an odd row, that might affect the odds of drawing from certain rows in subsequent picks. This dependency can break the independence between row-based and column-based events when drawing multiple times.
Single Pick vs. Multiple Picks: If we only draw one number in total, we’re effectively sampling with replacement in the sense that each cell is equally likely just once. But for multiple draws, the probabilities shift after each draw, which can introduce dependency between the events “odd row” and “first five columns” over successive draws.
How to handle it: For multiple draws, you can compute probabilities conditionally after each draw. Independence can hold for one draw but typically not across multiple draws without replacement.
Follow-up question 3: Could a partial reveal of information about the chosen number break independence?
Sometimes, you get partial information (for example, you learn “the chosen number is above a certain threshold” but not which row or column it came from).
Conditional Probability: Once you know the chosen number is above a threshold, the probability distribution might shift toward certain rows or columns if large values are clustered there.
Loss of Independence: Even if events were independent originally, conditioning on partial knowledge about the chosen value can induce dependence between row-based and column-based events. For instance, if most large numbers happen to be in odd rows or in the first five columns, learning “the chosen number is large” ties row and column information together.
How to handle it: This is a conditional probability situation: P(A∩B | partial info) vs. P(A | partial info) * P(B | partial info). One must re-check the independence using these conditional probabilities.
Follow-up question 4: How do real-world constraints on matrix structure create dependencies even under uniform cell selection?
In certain applications (e.g., images structured in patches, or sensor data arranged in grids), the matrix may have patterns:
Spatial Correlation: Adjacent rows or columns could store similar data (e.g., in image pixels or time-series). If the question is about some property that tends to occur in “clusters,” independence assumptions about row vs. column might fail.
Structured Layout: Real-world layouts often group related items in specific rows and columns. For example, if the first few columns store “low range” data and the last columns store “high range” data, then row-based properties (such as sensor type) might correlate with certain columns.
How to handle it: Investigate the matrix’s underlying structure and distribution of data. Even if you choose each cell uniformly, the content might be distributed in a way that leads to correlation between row-based events and column-based events.
Follow-up question 5: Is it possible for two events to be mutually exclusive yet also be “independent” in some contexts?
Ordinarily, if two events A and B are mutually exclusive (i.e., they cannot both happen at once), the only way for them to be independent is if one of them has probability 0. But in certain degenerate or tricky definitions:
Zero-Probability or Impossible Events: If P(A) = 0 or P(B) = 0, then P(A∩B)=0. By definition, 0 = 0 × anything, so they can appear “independent” in a trivial sense.
Pitfalls: Mutually exclusive events are usually not independent unless one is truly impossible. In typical finite matrix picks, you won’t see a case where A=“number from row 1” and B=“number from row 2” are independent, because that intersection is impossible yet both have nonzero probability on their own.
How to handle it: Recognize that true independence for nonzero events cannot happen if they are mutually exclusive. Always check for corner cases where one event might have zero probability.
Follow-up question 6: What if the matrix is extremely large, and we approximate probabilities with empirical frequencies?
For very large matrices (say thousands of rows and columns), it may be infeasible to compute exact probabilities directly by counting or summing. Instead, one might estimate frequencies from a sample:
Sampling Approach: Randomly pick a subset of cells and see how many fall into event A, event B, and the intersection. Then approximate probabilities by dividing the counts by the sample size.
Estimation Error and Confidence Intervals: With finite sampling, you get an estimate of P(A∩B) and of P(A)P(B), but you need to account for statistical variation. Independence would be suggested if these estimates are numerically close, but they might differ due to random sampling error.
Pitfalls: If the matrix distribution has hidden patterns (clustering, gradients, etc.), a naive sampling strategy might miss them, leading to incorrect independence conclusions.
How to handle it: Use stratified or systematic sampling to ensure coverage across rows and columns. Combine the sampling results with confidence intervals to draw robust conclusions about independence.
Follow-up question 7: How might independence assumptions break down if we modify event definitions?
Sometimes, you redefine events in more complex ways. For instance, event A could mean “the chosen number is in one of the odd-numbered rows and is greater than some value,” whereas event B might remain “the chosen number is in the first five columns.”
Complex Overlap: When events incorporate additional conditions (like a numerical threshold), the probability distribution of the chosen number can change drastically across different rows or columns, affecting independence.
Boundary Cases: Suppose large numbers predominantly appear in certain columns. Then an event restricting to large numbers is not independent of the event “in first five columns.”
How to handle it: Recalculate probabilities carefully for the newly defined events. Independence can’t be assumed to hold just because it held for simpler row-based vs. column-based definitions.
Follow-up question 8: What if the rows or columns represent categories with different inherent probabilities, but we only know aggregated counts?
Sometimes, each row represents a distinct category (e.g., different machine types) and each column a distinct feature range (e.g., low, medium, high), but the data is only available in aggregated form.
Aggregated Data: If all you have is the total count of how many items fall in each row category and each column category, you might not have the counts for each row-column combination. Without that, you can’t conclusively test P(A∩B)=P(A)P(B).
Risk of Misinterpretation: You might incorrectly assume independence if the marginal counts look balanced but the distribution across row-column intersections is skewed. Simpson’s paradox is a classic example where aggregated data can mask dependencies present in finer-grained data.
How to handle it: You must obtain or estimate the joint distribution of row-column intersections. If that’s missing, you’ll need additional assumptions or external data to test independence reliably.
Follow-up question 9: How do we handle “almost” independence (where P(A∩B) is close, but not exactly equal, to P(A)P(B))?
In real-life data, exact independence is rare. Instead, we might look for approximate independence or measure how far we are from perfect equality.
Statistical Measures: One might compute absolute difference |P(A∩B) – P(A)P(B)| or relative difference. A small difference might be considered negligible.
Significance Testing: For large data sets, even a tiny deviation might be statistically significant. One might use a chi-squared test or other hypothesis tests for independence in contingency tables.
Practical Implications: If near independence is good enough (e.g., for simplifying a large statistical model), then approximate independence might suffice. But in high-stakes contexts (like medical diagnoses), even small deviations might matter.
How to handle it: Define thresholds for approximate independence. Use robust statistical tests and consider whether small deviations are practically meaningful in the application context.
Follow-up question 10: Could we extend the concept of independence to more than two events involving rows and columns?
Yes. In higher-dimensional scenarios—say, multiple row-based events (odd row, prime row index, etc.) and multiple column-based events (first five columns, column indices are multiples of 2, etc.)—we might wonder if they’re mutually independent as a group.
Pairwise vs. Mutual Independence: Even if each pair of events is independent, the entire set of events might not be mutually independent. Mutual independence requires that for any subset of events, P of the intersection = product of the individual probabilities in that subset.
Complex Overlaps: If rows and columns define further subevents, you may get interactions (like a row event being correlated with a second row event in a way that only shows up when combined with a certain column event).
How to handle it: For mutual independence, systematically verify that all combinations adhere to P(intersection) = product of marginals. This quickly becomes complicated, so typically, we look at pairwise or simpler forms of independence unless we have strong reasons to test the entire set.