Commuter Ride-Choice Probability: Committed and Permanent Riders
Company: Lyft
Role: Data Scientist
Category: Statistics & Math
Difficulty: medium
Interview Round: Technical Screen
A ride-hailing company is modeling the daily behavior of commuters. Consider a population of $N$ commuters. Each commuter takes exactly **two trips per day**: a morning trip **to work** and an evening trip **back home**. For each trip the commuter independently decides whether to take a ride with the company:
- Probability of choosing the company for the **morning (to-work)** trip is $p$, where $0 < p < \tfrac{1}{3}$.
- Probability of choosing the company for the **evening (back-home)** trip is $2p$.
- The two trip decisions on the same day are independent of each other, and decisions are independent across days and across commuters.
A commuter is called a **"committed" rider on a given day** if they choose the company for **both** of that day's trips (morning and evening). Assume the evening trip is the more expensive of the two (it falls in the higher-priced evening peak), so taking the evening ride counts as taking a **high-priced ride** that day, while the morning ride is standard-priced.
Answer each of the following. Express every answer in terms of $p$ (and $N$ where relevant), and simplify.
**Part (a)** What is the probability that a randomly chosen commuter takes **at least one** ride with the company on a given day?
**Part (b)** What is the probability that a commuter becomes a **committed rider** (takes the company for both trips) on a given day?
**Part (c)** Consider a brand-new rider on **Day 1**. What is the probability that this rider takes **at least one** ride with the company again on **Day 2**? (Day 2 behavior follows the same model and is independent of Day 1.)
**Part (d)** Define becoming a **"permanent" rider within two days** as being a committed rider (both trips) on **at least one** of Day 1 or Day 2. What is the probability that the new rider becomes a permanent rider within two days?
**Part (e)** Given that a rider took **at least one** ride with the company on Day 2, what is the probability that they took **no high-priced (evening) ride on Day 1**?
```hint Set up the per-trip events
For a single day, let $W$ be the event "took the morning (to-work) ride," with $P(W)=p$, and $H$ the event "took the evening (high-priced, back-home) ride," with $P(H)=2p$. $W$ and $H$ are independent. Every quantity below is built from these two events and the across-day independence.
```
```hint At-least-one via the complement
"At least one ride in a day" is easiest through the complement: $1 - P(\text{neither trip}) = 1 - (1-p)(1-2p)$. Reuse this single per-day probability — call it $q$ — wherever a part asks about a full day or a second day.
```
```hint Across-day and conditioning
Days are independent, so "within two days" / "again on Day 2" multiply or complement per-day probabilities. For Part (e), the conditioning event (a ride on Day 2) is independent of Day 1, so Bayes collapses: condition only re-weights, and the Day-1 question becomes a marginal about Day 1 alone.
```
### Constraints & Assumptions
- Within a day, the morning and evening trip decisions are independent; $P(\text{morning})=p$, $P(\text{evening})=2p$.
- The constraint $p<\tfrac13$ keeps $2p<\tfrac23<1$, so all probabilities are valid.
- Decisions are independent across days and across the $N$ commuters; each day follows the identical model.
- "High-priced ride" refers exactly to the evening (back-home) trip.
### Clarifying Questions to Ask
- Are the morning and evening decisions independent, or can a commuter's morning choice change their evening probability?
- Does "at least one ride" mean either trip, and does "committed/both" require the *same* day?
- Are Day 1 and Day 2 independent and identically distributed under this model?
- For the conditional in Part (e), is the Day-2 ride event independent of all Day-1 events?
- Is the evening trip always the high-priced one, and the morning trip never high-priced?
### What a Strong Answer Covers
- Correctly identifies the two per-trip events and their probabilities and uses **independence** to multiply.
- Uses the **complement** for "at least one" rather than summing overlapping cases.
- Recognizes **across-day independence** and applies it cleanly to Parts (c) and (d).
- For Part (e), recognizes the conditioning event is **independent** of Day 1, so the conditional probability reduces to a Day-1 marginal — and does not blindly grind through Bayes.
- Keeps the algebra tidy, factors where possible, and sanity-checks that every result lies in $[0,1]$ given $p<\tfrac13$.
### Follow-up Questions
- How would the answers to Parts (a) and (b) change if the morning and evening decisions were **positively correlated** (a commuter who rides in the morning is more likely to ride home)?
- Across $N$ independent commuters, what is the distribution of the number who become committed on a given day, and its mean and variance?
- If you observed the empirical fraction of committed riders in real data, how would you estimate $p$ and put a confidence interval on it?
- Generalize Part (d): what is the probability of becoming a permanent rider within $k$ days, and what does it converge to as $k\to\infty$?
Quick Answer: This question tests a data scientist's ability to apply probability theory to multi-event, multi-day behavioral models, covering independence, complementary counting, and conditional probability. It evaluates practical fluency with discrete probability calculations, including Bayes' theorem and compound event reasoning, commonly assessed in quantitative data science interviews.