Meta·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Meta data scientist interview with a probability and statistics focus. Two questions, both conditional probability flavored, one a binomial conditioning problem and the other a Bayesian setup with a three-state room model. Pretty math-heavy for what I expected.

Questions Asked (2)

There are N conference rooms and K meetings assigned uniformly at random. Given that room 1 is not empty, what is the expected number of meetings in room 1?

Algorithms & Data StructuresProduct Analytics & Metrics

Author's notes

The binomial part wasn't too bad once I remembered you're conditioning on at least one meeting landing there.

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AI HintsAI Generated

Suggested Approach

First, clarify that each meeting is assigned independently and uniformly to one of N rooms, so the number of meetings in room 1 follows a Binomial(K, 1/N) distribution. Then compute the conditional expectation E[X | X > 0] using the formula E[X] / P(X > 0), where X is the number of meetings in room 1.

Pro tip: Show that you can derive the result from first principles and then sanity-check with extreme cases (e.g., K=1, N large) to demonstrate practical intuition. Mention that in real-world settings, uniform assignment is often unrealistic, and you'd validate with data.

1. Define the random variable

Let X be the number of meetings in room 1. State that X ~ Binomial(K, 1/N) under the uniform random assignment assumption.

2. Compute unconditional expectation

Calculate E[X] = K/N. This is the expected number of meetings in room 1 without any conditioning.

3. Compute probability room 1 is not empty

Find P(X > 0) = 1 - P(X = 0) = 1 - (1 - 1/N)^K.

4. Apply conditional expectation formula

Use E[X | X > 0] = E[X] / P(X > 0) = (K/N) / (1 - (1 - 1/N)^K). Simplify if needed.

5. Sanity check and interpret

Verify with edge cases (e.g., K=1 gives 1, K large gives K/N) and discuss implications for capacity planning.

Key Points to Mention

Binomial distribution for the number of meetings in a given room
Conditional expectation formula: E[X | A] = E[X] / P(A) when A is a subset of {X > 0}
Probability that room 1 is empty: (1 - 1/N)^K
Assumption of independent uniform assignment
Edge cases: K=1, N=1, large K
Practical implications: room utilization, overbooking, and non-uniform real-world distributions

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

Two rooms exist where the three states (both occupied, exactly one occupied, both empty) are equally likely. If you pick a room at random without looking, what's the probability it's occupied? Then, given you enter and find it occupied, what's the probability the other room is also occupied?

Algorithms & Data StructuresA/B Testing & Experimentation

Author's notes

This is the part that got me.

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AI HintsAI Generated

Suggested Approach

First, clarify that the three states are equally likely (each probability 1/3). Then compute the marginal probability of a randomly chosen room being occupied by summing over states. For the conditional probability, apply Bayes' theorem or directly count the cases where the chosen room is occupied and the other is also occupied.

Pro tip: Explicitly state the assumption that the three states are equally likely and that the room selection is uniform and independent of the state. This avoids ambiguity and shows rigorous thinking.

1. Define the states and probabilities

Let the states be: A = both occupied, B = exactly one occupied, C = both empty. Each has probability 1/3.

2. Compute marginal probability of a room being occupied

For a randomly chosen room, P(occupied) = P(occupied|A)*P(A) + P(occupied|B)*P(B) + P(occupied|C)*P(C) = 1*(1/3) + 0.5*(1/3) + 0*(1/3) = 1/2.

3. Set up conditional probability

We need P(other occupied | chosen occupied). Use Bayes: P(other occupied | chosen occupied) = P(both occupied) / P(chosen occupied).

4. Calculate the conditional probability

P(both occupied) = 1/3. P(chosen occupied) = 1/2. So the conditional probability = (1/3) / (1/2) = 2/3.

5. Interpret and sanity-check

The result 2/3 makes sense: given you see an occupied room, it's more likely to come from the both-occupied state. Verify by enumerating equally likely scenarios if needed.

Key Points to Mention

Equally likely states: each has probability 1/3.
Law of total probability to find marginal P(occupied) = 1/2.
Bayes' theorem or conditional probability formula.
P(both occupied) = 1/3, P(chosen occupied) = 1/2, so conditional = 2/3.
Independence assumption: room selection is random and independent of occupancy state.
Potential for confusion: the answer is not 1/2 because observing an occupied room provides information about the state.

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