SIG (Susquehanna)·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Probability question from a SIG data scientist interview. Classic Bayesian setup, nothing too wild, but the arithmetic is easy to fumble under pressure.

Questions Asked (1)

Factory A produces 40% red and 60% black widgets; Factory B produces 80% red and 20% black widgets. A factory is chosen at random, then two widgets are drawn from it. Given that both widgets are black, what is the probability they came from Factory A?

Product Analytics & MetricsAlgorithms & Data Structures

Author's notes

This is straight Bayes.

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AI HintsAI Generated

Suggested Approach

Recognize this as a Bayes' theorem problem with a twist: the two draws are without replacement, so the probability of two black widgets depends on the factory's composition. Compute the likelihood of drawing two black widgets from each factory, then apply Bayes' theorem to find the posterior probability of Factory A given the evidence.

Pro tip: Always clarify whether the draws are with or without replacement—here, 'two widgets are drawn' implies without replacement, which significantly changes the likelihoods. Mentioning this assumption upfront shows attention to detail and avoids a common pitfall.

1. Define events and priors

Let A be the event that Factory A is chosen, B be Factory B. Priors: P(A)=0.4, P(B)=0.6. Let E be the event that both widgets drawn are black.

2. Compute likelihoods

Calculate P(E|A) and P(E|B) assuming draws without replacement. For Factory A: (0.6)*(0.6) = 0.36? Wait, without replacement: (0.6)*(0.6) is incorrect; need to account for finite population. Actually, if we assume large population, it's approximately (0.6)^2 = 0.36. But if we consider exact without replacement from a finite batch, we need the total number of widgets. Since not given, assume large population or with replacement? The problem says 'two widgets are drawn from it'—typically implies without replacement, but without total count, we approximate as independent. So P(E|A) ≈ 0.6 * 0.6 = 0.36. Similarly, P(E|B) ≈ 0.2 * 0.2 = 0.04. However, if we strictly consider without replacement and assume a large population, it's fine. But if we assume a small batch, we need more info. So state assumption: assume large population or with replacement for simplicity. Alternatively, if we assume exactly two widgets drawn without replacement from an infinite population, it's the same as with replacement. So proceed with 0.36 and 0.04.

3. Apply Bayes' theorem

4. Sanity check and interpret

Verify that the result makes sense: Factory B is more likely to produce black widgets, but drawing two black widgets is much more likely from Factory A (0.36 vs 0.04), so the posterior shifts heavily toward Factory A despite its lower prior. The answer is approximately 85.7%.

Key Points to Mention

Bayes' theorem and its application to conditional probability
Prior probabilities: P(A)=0.4, P(B)=0.6
Likelihoods: P(2 black | A) and P(2 black | B)
Assumption of independent draws (with replacement or large population)
Calculation of the denominator as total probability of evidence
Interpretation of the posterior probability in context

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.