MathWorks·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

MathWorks software engineer interview that leaned heavily on probability and combinatorics. The math wasn't brutal but you'd better show your work clearly or it gets awkward fast.

Questions Asked (1)

A bag has 6 red balls and 4 blue balls. You draw 3 without replacement. What's the probability that at least 2 of the drawn balls are red? Walk through your calculations.

Algorithms & Data Structures

Author's notes

The 'at least two' phrasing is where I slowed down.

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AI HintsAI Generated

Suggested Approach

Recognize this as a hypergeometric probability problem and compute the probability of exactly 2 red balls and exactly 3 red balls, then sum them. Alternatively, use the complement: 1 minus the probability of 0 or 1 red balls. Clearly state the combinatorial formula and show the arithmetic.

Pro tip: After calculating, sanity-check the result by considering the expected number of red balls (1.8) and the symmetry of the distribution; a probability around 0.67 is reasonable. Also, mention that this is equivalent to the hypergeometric distribution, which is useful in many engineering contexts.

1. Define the problem and identify the distribution

State that you are drawing without replacement from a finite population, so the number of red balls follows a hypergeometric distribution. Define the parameters: population size N=10, successes K=6, draws n=3.

2. Determine the required events

The event 'at least 2 red balls' means exactly 2 red or exactly 3 red. These are mutually exclusive, so you can sum their probabilities.

3. Compute probabilities using combinations

Calculate P(2 red) = C(6,2)*C(4,1)/C(10,3) and P(3 red) = C(6,3)*C(4,0)/C(10,3). Show the combination values: C(6,2)=15, C(4,1)=4, C(10,3)=120, C(6,3)=20, C(4,0)=1.

4. Sum the probabilities and simplify

Add the two probabilities: (60/120) + (20/120) = 80/120 = 2/3 ≈ 0.6667. Present the final answer clearly.

5. Verify with complement (optional)

As a check, compute P(0 red) = C(6,0)*C(4,3)/120 = 4/120 and P(1 red) = C(6,1)*C(4,2)/120 = 36/120. Sum = 40/120, so 1 - 40/120 = 80/120 = 2/3, confirming the result.

Key Points to Mention

Drawing without replacement implies dependence, so use combinations (hypergeometric) rather than binomial.
The events 'exactly 2 red' and 'exactly 3 red' are mutually exclusive, so probabilities add.
Combination formula: C(n, k) = n! / (k! (n-k)!).
Total number of ways to draw 3 balls from 10 is C(10,3) = 120.
The complement method (1 - P(0 or 1 red)) is a valid alternative and can serve as a check.
The final probability is 2/3 or approximately 66.67%.

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