Recognize this as a Bayesian inference problem: you need P(Factory A | two black widgets). Use Bayes' theorem with the prior probability of choosing each factory (0.5) and the likelihood of drawing two black widgets from each factory (computed via the binomial distribution). Compute the posterior probability by multiplying prior and likelihood for each factory, then normalizing.
Pro tip: After computing the exact probability, mention that in an A/B testing context, this is analogous to updating your belief about which variant is better given observed data. Also, sanity-check the result: since Factory B has a lower probability of black widgets, observing two black widgets should decrease the probability it's Factory A, so the posterior should be less than 0.5.
State that you need P(A | BB). Note the prior P(A)=0.5, P(B)=0.5, and the conditional probabilities of drawing a black widget from each factory: P(black|A)=0.6, P(black|B)=0.2.
Assuming independent draws with replacement (or from a large population), calculate P(BB|A) = 0.6 * 0.6 = 0.36 and P(BB|B) = 0.2 * 0.2 = 0.04.
Compute the unnormalized posteriors: P(A)*P(BB|A) = 0.5 * 0.36 = 0.18 and P(B)*P(BB|B) = 0.5 * 0.04 = 0.02. Then normalize: P(A|BB) = 0.18 / (0.18 + 0.02) = 0.9.
Explain that there is a 90% chance the widgets came from Factory A. Relate this to A/B testing: updating prior beliefs with observed data to make decisions.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.