Aspiring data scientists often face challenging interview questions at top tech companies like Microsoft, Meta (Facebook), and Lyft. These questions are designed to assess your understanding of probability, statistics, experimental design, algorithmic thinking, and your ability to communicate complex ideas. In this article, we’ll break down several real-world and theoretical interview questions, explain the concepts involved, and provide step-by-step solutions so you can ace your data science interviews.

Data Scientist Interview Questions - Microsoft and Meta

1. Coin Flipping: HHT vs HTT — Which Comes First?

Question

You're given a fair coin. You flip the coin until either Heads Heads Tails (HHT) or Heads Tails Tails (HTT) appears. Is one more likely to appear first? If so, which one and with what probability?

Concepts Involved

• Markov Chains
• Conditional Probability
• Pattern Occurrence in Random Sequences

Solution

Let’s define the two sequences as targets: HHT and HTT. We want to know, if we repeatedly flip a fair coin, which sequence is more likely to appear first — or if their chances are equal.

Let's model this as a Markov process, keeping track of the "history" of the last one or two flips to know how close we are to either pattern.

Defining the States

• S0: Start state, no flips yet, or last flip was 'T'.
• S1: Last flip was 'H'.
• S2: Last two flips were 'HH'.
• S3: Last two flips were 'HT'.
• SHHT: Sequence HHT appeared, stop.
• SHTT: Sequence HTT appeared, stop.

Defining Probabilities

Let’s define:

• \( P_A \): Probability that HHT occurs before HTT, starting from S0.
• \( P_B \): Probability that HHT occurs before HTT, starting from S1 (last is H).
• \( P_C \): Probability that HHT occurs before HTT, starting from S2 (last two flips are HH).
• \( P_D \): Probability that HHT occurs before HTT, starting from S3 (last two flips are HT).

Writing the Equations

Let's build the system by considering all possible transitions:

• S0:
  • Flip H: move to S1 (\( \frac{1}{2} \) probability)
  • Flip T: stay at S0 (\( \frac{1}{2} \) probability)
• S1:
  • Flip H: move to S2 (\( \frac{1}{2} \))
  • Flip T: move to S3 (\( \frac{1}{2} \))
• S2:
  • Flip T: HHT is formed (\( \frac{1}{2} \)), stop.
  • Flip H: stay at S2 (\( \frac{1}{2} \)), sequence is still HH.
• S3:
  • Flip T: HTT is formed (\( \frac{1}{2} \)), stop.
  • Flip H: move to S1 (\( \frac{1}{2} \)), sequence is H.

Expressing as equations:

• \( P_A = \frac{1}{2} P_B + \frac{1}{2} P_A \)
• \( P_B = \frac{1}{2} P_C + \frac{1}{2} P_D \)
• \( P_C = \frac{1}{2} \times 1 + \frac{1}{2} P_C \)
• \( P_D = \frac{1}{2} \times 0 + \frac{1}{2} P_B \)

Let’s solve them step by step.

Solving the Equations

1. Solve for \( P_C \):
   \( P_C = \frac{1}{2} \times 1 + \frac{1}{2} P_C \)
   \( P_C - \frac{1}{2} P_C = \frac{1}{2} \)
   \( \frac{1}{2} P_C = \frac{1}{2} \)
   \( P_C = 1 \)
2. Solve for \( P_D \):
   \( P_D = \frac{1}{2} \times 0 + \frac{1}{2} P_B = \frac{1}{2} P_B \)
3. Plug into \( P_B \):
   \( P_B = \frac{1}{2} P_C + \frac{1}{2} P_D \)
   \( P_B = \frac{1}{2} \times 1 + \frac{1}{2} \times \frac{1}{2} P_B \)
   \( P_B = \frac{1}{2} + \frac{1}{4} P_B \)
   \( P_B - \frac{1}{4} P_B = \frac{1}{2} \)
   \( \frac{3}{4} P_B = \frac{1}{2} \)
   \( P_B = \frac{2}{3} \)
4. Now, \( P_D = \frac{1}{2} P_B = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \)
5. Finally, \( P_A = \frac{1}{2} P_B + \frac{1}{2} P_A \)
   \( P_A - \frac{1}{2} P_A = \frac{1}{2} P_B \)
   \( \frac{1}{2} P_A = \frac{1}{2} \times \frac{2}{3} \)
   \( P_A = \frac{2}{3} \)

Conclusion

HHT appears before HTT with probability \( \frac{2}{3} \), and HTT appears first with probability \( \frac{1}{3} \). Though the sequences look similar, the difference in overlap makes HHT more likely to appear first.

2. Lyft 50% Rider Discount Promotion: Experiment Design and Metrics

Question

You work as a data scientist for Lyft. A VP asks how you would evaluate whether a 50% rider discount promotion is a good or bad idea? How would you implement it? What metrics would you track?

Concepts Involved

• Experimentation and A/B Testing
• Key Performance Indicators (KPIs)
• Statistical Testing and Causal Inference
• Revenue and Customer Lifetime Value (CLV) Analysis

Solution

Step 1: Define the Objective

First, clarify what “good” means. Are we trying to maximize revenue, increase active users, improve retention, or enter a new market? For this example, let’s assume the goal is to grow active riders and long-term revenue.

Step 2: Design the Experiment

• Randomly split similar users into two groups: Treatment (gets 50% off) and Control (no discount).
• Ensure stratification on key dimensions (location, recent activity, demographics) to avoid imbalances.
• Determine sample size using power analysis to detect a meaningful difference with sufficient confidence (typically, 80% power at 5% significance).
• Run the promotion for a defined period (e.g., 4 weeks).

Step 3: Key Metrics to Track

Step 4: Analyze the Results

• Use difference-in-means statistical tests (e.g., t-test, Mann-Whitney U) to compare metrics between treatment and control.
• Adjust for confounding variables using regression or matching if needed.
• Monitor for “adverse selection” (are power users disproportionately using the discount?).
• Check for “cannibalization” (are users who would have ridden anyway just paying less?).

Step 5: Make Recommendations

• If increased lifetime value and retention outweighs short-term profit loss, the promo may be “good.”
• If promo attracts only price-sensitive users who churn post-promo, it may be “bad.”
• Suggest further segmentation (e.g., offer only to new users or low-frequency riders).

Sample Implementation (Pseudocode)

import numpy as np
from scipy.stats import ttest_ind

# Assume bookings_treatment and bookings_control are numpy arrays
t_stat, p_val = ttest_ind(bookings_treatment, bookings_control, equal_var=False)
if p_val < 0.05:
    print("Statistically significant difference in gross bookings!")
else:
    print("No significant difference found.")

3. Bayesian Probability: Friends Lying About Seattle Rain

Question

You are about to get on a plane to Seattle. You want to know if you should bring an umbrella. You call 3 random friends of yours who live there and ask each independently if it's raining. Each of your friends has a 2/3 chance of telling you the truth and a 1/3 chance of messing with you by lying. All 3 friends tell you that "Yes" it is raining. What is the probability that it's actually raining in Seattle?

Concepts Involved

• Bayes’ Theorem
• Conditional Probability
• Independence of Events

Solution

Let’s define:

• R: It is raining in Seattle.
• NR: It is NOT raining in Seattle.
• Each friend independently tells the truth with probability \( \frac{2}{3} \), lies with \( \frac{1}{3} \).
• All 3 friends say "YES".

Let’s Calculate Probabilities

By Bayes’ Theorem:

\[ P(R | \text{all say "YES"}) = \frac{P(\text{all say "YES"} | R) P(R)}{P(\text{all say "YES"})} \]

Assume prior \( P(R) = P(NR) = 0.5 \) (if nothing else is known).

If it is raining (\( R \)):

• Each friend says "YES" with probability \( \frac{2}{3} \) (truth) + \( \frac{1}{3} \) (lying, but since it's raining, lying would mean saying "NO"). So, only truth-tellers say "YES".

Therefore, \[ P(\text{Friend says "YES"} | R) = \frac{2}{3} \] So, all three say "YES": \[ P(\text{all "YES"} | R) = \left( \frac{2}{3} \right)^3 = \frac{8}{27} \]

If it is NOT raining (\( NR \)):

• Each friend tells the truth (\( \frac{2}{3} \)): says "NO".
• Each friend lies (\( \frac{1}{3} \)): says "YES".

So, \[ P(\text{Friend says "YES"} | NR) = \frac{1}{3} \] \[ P(\text{all "YES"} | NR) = \left( \frac{1}{3} \right)^3 = \frac{1}{27} \]

Now, \[ P(\text{all "YES"}) = P(\text{all "YES"} | R) \times P(R) + P(\text{all "YES"} | NR) \times P(NR) \] \[ = \frac{8}{27} \times 0.5 + \frac{1}{27} \times 0.5 = \frac{8 + 1}{27} \times 0.5 = \frac{9}{27} \times 0.5 = \frac{1}{3} \times 0.5 = \frac{1}{6} \]

So, \[ P(R | \text{all "YES"}) = \frac{\frac{8}{27} \times 0.5}{\frac{\[ P(R | \text{all "YES"}) = \frac{\frac{8}{27} \times 0.5}{\frac{1}{6}} \] Let's calculate numerator and denominator: - Numerator: \(\frac{8}{27} \times 0.5 = \frac{8}{54} = \frac{4}{27}\) - Denominator: \(\frac{1}{6}\) Therefore, \[ P(R | \text{all "YES"}) = \frac{\frac{4}{27}}{\frac{1}{6}} = \frac{4}{27} \times \frac{6}{1} = \frac{24}{27} = \frac{8}{9} \]

Interpretation

Even though your friends are not always reliable, if all three independently say “YES, it’s raining,” the probability it is actually raining in Seattle is \(\frac{8}{9}\) (about 88.9%).

This is a great example of Bayesian updating: even with imperfect sources, consensus among them can still provide strong evidence for the truth.

4. Text Justification (Microsoft): Evenly Distribute Spaces

Question

Given an array of words and a maxWidth parameter, format the text so that each line has exactly maxWidth characters. Pad extra spaces when necessary so that each line has exactly maxWidth characters. Extra spaces between words should be distributed as evenly as possible. If spaces don't divide evenly, the leftmost slots get more spaces than those on the right.

Concepts Involved

• Greedy Algorithms
• String Manipulation
• Edge Case Handling

Solution Overview

This is a classic text justification problem. The approach is:

• Group words into lines so the total length (including at least one space between words) does not exceed maxWidth.
• For each line (except the last), distribute the spaces as evenly as possible between words. If spaces don’t divide evenly, assign more to the leftmost slots.
• The last line should be left-justified (words separated by one space, pad the right).

Algorithm Steps

1. Line Construction: Iterate through the words, greedily adding as many as possible to the current line without exceeding maxWidth.
2. Space Distribution:
   • If it's the last line or the line has only one word: left-justify (words separated by single spaces, pad the rest on the right).
   • Otherwise: distribute spaces evenly between words. If not even, assign the extra to the left slots.

Python Implementation

def fullJustify(words, maxWidth):
    res, cur, num_of_letters = [], [], 0
    for w in words:
        if num_of_letters + len(w) + len(cur) > maxWidth:
            # Time to justify cur
            spaces = maxWidth - num_of_letters
            if len(cur) == 1:
                # Only one word, left-justify
                res.append(cur[0] + ' ' * spaces)
            else:
                space_between = spaces // (len(cur) - 1)
                extra = spaces % (len(cur) - 1)
                line = ''
                for i in range(len(cur)):
                    line += cur[i]
                    if i < len(cur) - 1:
                        # Extra spaces to leftmost slots
                        line += ' ' * (space_between + (1 if i < extra else 0))
                res.append(line)
            cur, num_of_letters = [], 0
        cur += [w]
        num_of_letters += len(w)
    # Last line
    last_line = ' '.join(cur)
    last_line += ' ' * (maxWidth - len(last_line))
    res.append(last_line)
    return res

# Example usage:
words = ["This", "is", "an", "example", "of", "text", "justification."]
maxWidth = 16
justified = fullJustify(words, maxWidth)
for line in justified:
    print(f"'{line}'")

Explanation

• We build each line by adding words until adding another would exceed maxWidth (including spaces).
• Spaces are distributed as evenly as possible. If there are extra spaces, they are assigned to the leftmost gaps first (per the problem statement).
• The last line is left-justified, with single spaces between words and any extra space at the end.

Sample Output

Conclusion

Data science interviews at Microsoft, Meta, Lyft, and similar companies challenge you to combine probability, experimentation, and algorithmic thinking. For probability and Bayes’ rule questions, take time to define your events and use clear step-by-step calculations. For experiment design (like Lyft’s discount), always clarify objectives, design robust experiments, and choose meaningful metrics. For algorithmic or coding questions, break down the problem, consider edge cases, and write clean, well-commented code. With practice on these types of questions, you’ll be well-prepared to impress in your next data science interview.

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