Sig·Software Engineer·Technical Phone Screen·Junior

Junior

Jun 2026

Summary

SIG quant researcher interview with a probability problem that looks clean on the surface but has a few ways to go sideways if you're not careful about how you set up the distribution.

Questions Asked (1)

In a bakery, each customer independently buys a croissant with probability 0.8 or a muffin with probability 0.2. There are only 2 croissants left and 5 customers in line. What is the probability that 2 croissants are enough to cover everyone who wants one?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

So this is basically asking for P(X <= 2) where X is binomial with n=5 and p=0.8.

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AI HintsAI Generated

Suggested Approach

Recognize this as a binomial probability problem where each customer independently wants a croissant with probability 0.8. The event that 2 croissants are enough is equivalent to at most 2 customers out of 5 wanting a croissant. Compute P(X ≤ 2) using the binomial formula or by summing individual probabilities for X=0,1,2.

Pro tip: After computing the probability, briefly mention that this is a classic binomial distribution application and that the same logic extends to inventory planning for any number of items and customers. This shows you can generalize and apply the concept to real-world engineering problems.

1. Identify the distribution

Define X as the number of customers who want a croissant. X follows a binomial distribution with n=5 and p=0.8.

2. Define the success condition

The condition '2 croissants are enough' means X ≤ 2. So we need to find P(X ≤ 2).

3. Compute individual probabilities

Calculate P(X=0), P(X=1), and P(X=2) using the binomial formula: P(X=k) = C(n,k) * p^k * (1-p)^(n-k).

4. Sum the probabilities

Add the probabilities from step 3 to get P(X ≤ 2).

5. Interpret and verify

State the final probability and sanity-check that it is reasonable (e.g., since p=0.8, the probability should be low).

Key Points to Mention

Binomial distribution assumptions: independent trials, constant probability, fixed number of trials.
The complement approach: P(X ≤ 2) = 1 - P(X ≥ 3) could also be used, but direct computation is simpler here.
Exact calculation: P(X=0)=0.00032, P(X=1)=0.0064, P(X=2)=0.0512, sum = 0.05792.
Real-world implication: This low probability suggests that having only 2 croissants is insufficient most of the time.
Generalization: The same method applies to any inventory problem with independent demand.
Potential trade-offs: In a real system, you might consider overstocking vs. understocking costs, which relates to engineering trade-offs.

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