The binomial formula itself was fine but I fumbled the edge cases a bit.
Recognize that the number of users linking wallets follows a Binomial(n, p) distribution. Then derive the probability of at least one link as 1 - (1-p)^n and the probability of exactly k links using the binomial PMF. Discuss edge cases and the Poisson approximation for large n and small p.
Pro tip: Connect the math to Coinbase's business context by mentioning how these probabilities inform growth metrics, such as conversion rates and expected active users, and how the Poisson approximation can simplify modeling for large user bases with low link rates.
Let X be the number of users who link their wallet. State that X ~ Binomial(n, p) since each user links independently with probability p.
Use the complement rule: P(X ≥ 1) = 1 - P(X = 0) = 1 - (1-p)^n.
Use the binomial PMF: P(X = k) = C(n, k) * p^k * (1-p)^(n-k), for k = 0, 1, ..., n.
For p=0, P(X ≥ 1)=0 and P(X=k)=1 if k=0 else 0. For p=1, P(X ≥ 1)=1 and P(X=k)=1 if k=n else 0. For large n, discuss behavior and Poisson approximation when p is small.
When n is large and p is small such that λ = np is moderate, X approximately follows Poisson(λ). Then P(X=k) ≈ e^{-λ} λ^k / k! and P(X ≥ 1) ≈ 1 - e^{-λ}.
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Linearity of expectation saves you here and I said that immediately, which felt good.
Start by defining the metric precisely: the expected number of users who link a wallet, which is the sum of individual link probabilities. Then model the probability for each user using a funnel (e.g., signup → wallet creation → link) and explicitly state where independence is assumed (across users) and where it is not required (within a user's funnel steps). Finally, discuss how to estimate the parameters from data and validate the model.
Pro tip: Emphasize that independence across users is sufficient for linearity of expectation, but within-user dependencies (e.g., a user's steps) must be handled via conditional probabilities or joint distributions. This shows you understand the subtlety beyond just multiplying probabilities.
Clarify what 'link a wallet' means (e.g., connecting an external wallet to a Coinbase account) and the population of users (e.g., new signups in a given period). Specify the time frame and any relevant segments.
For each user, model the probability of linking as a product of conditional probabilities for sequential steps (e.g., P(link) = P(signup) * P(create wallet | signup) * P(link | create wallet)). Acknowledge that these steps are not independent, so use conditional probabilities.
The expected number of users who link is the sum of individual link probabilities. This requires independence across users only for variance calculations, but the expectation itself does not require independence across users.
Use historical data to estimate the conditional probabilities, possibly with cohort analysis or survival models. Validate the model by comparing predicted vs. actual link counts and checking for overdispersion.
Explicitly state assumptions: independence across users (for variance), no interference, and stationarity. Discuss extensions like heterogeneous user segments, time-varying probabilities, or network effects.
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Start by setting up the problem: let p be the probability a user links, n the number of users, and assume independence. Then use the Poisson approximation with λ = np to estimate P(at least one link) ≈ 1 - e^{-λ}. For the error term, compare the exact binomial probability 1 - (1-p)^n with the Poisson approximation and expand to first order in p, showing that the Poisson approximation underestimates the true probability by approximately λp/2.
Pro tip: Mention that the error direction (underestimation) is consistent with the fact that the Poisson approximation ignores the negative binomial correlation, and that in practice for Coinbase-scale data (large n, small p), the error is negligible unless p is not truly small.
Let p be the probability that a single user links, n the number of users, and assume independent linking events. Define λ = np as the expected number of links.
For small p and large n, the number of links approximately follows a Poisson distribution with mean λ. Thus, P(at least one link) ≈ 1 - e^{-λ}.
The exact probability of at least one link is 1 - (1-p)^n. Use this to compute the error of the Poisson approximation.
Expand both expressions to first order in p (or λ/n) and subtract. Show that the error is approximately λp/2, meaning the Poisson approximation underestimates the true probability.
Conclude that the Poisson approximation is an underestimate, with error proportional to p. For small p, the error is negligible, but it grows with p.
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Straightforward plug-and-chug after the theory parts.
Recognize this as a binomial distribution problem with n=5 and p=0.30. Compute P(at least one) using the complement rule, P(exactly 2) using the binomial PMF, and the expected value as n*p. Show your work clearly and interpret the results in the context of user behavior.
Pro tip: After computing the numbers, briefly explain what they mean for the business—e.g., 'On average, 1.5 out of 5 users link, but there's a 16.8% chance nobody links, which could impact engagement metrics.' This demonstrates you can translate math into actionable insights.
State that the number of users who link follows a binomial distribution with parameters n=5 and p=0.30, assuming independence.
Use the complement rule: P(at least one) = 1 - P(none) = 1 - (0.70)^5. Calculate and round appropriately.
Apply the binomial PMF: P(X=2) = C(5,2) * (0.30)^2 * (0.70)^3. Compute the combination and powers.
Use the formula E[X] = n * p = 5 * 0.30 = 1.5. Mention that this is the long-run average.
Briefly explain what these probabilities and expected value mean in the context of user linking behavior, e.g., likelihood of at least one link, typical number of links.
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The conditional at the end is where people slip up and I nearly did.
First, clarify that A and B are independent events with given probabilities. Then apply the addition rule for P(A or B) and multiplication rule for P(A and B). Finally, use the definition of conditional probability to compute P(A and B | A or B).
Pro tip: In a data science interview, always state your assumptions (e.g., independence) and explain how you would validate them with data. This shows you think beyond the math and consider real-world implications.
Confirm that A and B are independent events, meaning P(A and B) = P(A) * P(B). Define the events clearly: A = user A links, B = user B links.
Since A and B are independent, multiply their probabilities: P(A and B) = 0.40 * 0.60 = 0.24.
Use the addition rule: P(A or B) = P(A) + P(B) - P(A and B) = 0.40 + 0.60 - 0.24 = 0.76.
Apply the conditional probability formula: P(A and B | A or B) = P((A and B) and (A or B)) / P(A or B). Since (A and B) implies (A or B), the numerator is P(A and B). Thus, 0.24 / 0.76 ≈ 0.3158.
Explain what these probabilities mean for Coinbase: e.g., the chance that at least one user links is 76%, both link is 24%, and given at least one links, the chance both link is about 31.6%.
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This was the hardest part and I definitely didn't nail it.
First, use the inclusion-exclusion principle to express P(A and B) in terms of the given probabilities, then apply the constraints that probabilities must be between 0 and 1 and that P(A and B) cannot exceed the marginals. Next, compute the conditional probability range by dividing the intersection range by P(A or B) = 0.70. Finally, interpret the boundaries in terms of dependence structures: maximum overlap (A subset of B) and minimum overlap (mutually exclusive with the union fixed).
Pro tip: Emphasize that the conditional probability is just a linear transformation of the intersection probability, so the range scales by 1/0.70. Also, note that the dependence structures correspond to extreme positive and negative dependence, which is crucial for understanding risk in financial data.
Use P(A or B) = P(A) + P(B) - P(A and B) to solve for P(A and B) = 0.40 + 0.60 - 0.70 = 0.30. This is the only possible value given the union and marginals, so the range is a single point.
Verify that 0.30 is valid: it must be ≤ min(0.40, 0.60) = 0.40 and ≥ max(0, 0.40+0.60-1) = 0. Since 0.30 satisfies these, it is feasible. Thus the range for P(A and B) is [0.30, 0.30].
Calculate P(A and B | A or B) = P(A and B) / P(A or B) = 0.30 / 0.70 ≈ 0.4286. Since the intersection is fixed, the conditional probability is also fixed at approximately 0.4286.
The single value corresponds to a specific dependence structure: since P(A and B) = 0.30, which is less than P(A)P(B)=0.24? Actually 0.30 > 0.24, so events are positively dependent. The structure is such that the overlap is exactly 0.30, which is the only possibility given the union and marginals.
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