Virtu Financial·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Interviewed for a Data Scientist role at Virtu Financial and got hit with a probability puzzle that looked deceptively clean on the surface. The green ball mechanic is what trips you up if you're not careful.

Questions Asked (1)

You have a bag with 5 red, 5 blue, and 5 green balls. You draw one ball at a time without replacement, except green balls get returned to the bag after each draw. You keep red and blue balls. Once you're holding 3 balls, what's the probability that at least 2 of them are red? Follow-up: same setup but now there are 10 red balls instead of 5. How does that change the answer?

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The green ball return mechanic is where I got tangled.

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AI HintsAI Generated

Suggested Approach

Clarify the process: each draw, if green, it's returned and does not count toward the 3 held balls; if red or blue, it's kept. So we are waiting for the first 3 non-green draws, and each such draw is red with probability 5/10 (or 10/15 in follow-up) independently? Actually, because greens are returned, the composition of non-green balls remains constant (5 red, 5 blue) until we have 3 kept balls. Thus the number of reds among the 3 kept balls follows a Binomial(3, 0.5) distribution. Then compute P(X >= 2) = P(X=2) + P(X=3) = 3*(0.5)^3 + (0.5)^3 = 4/8 = 0.5. For the follow-up, with 10 red and 5 blue, the probability of red on each non-green draw is 10/15 = 2/3, so X ~ Binomial(3, 2/3), and P(X >= 2) = 3*(2/3)^2*(1/3) + (2/3)^3 = 3*(4/9)*(1/3) + 8/27 = 12/27 + 8/27 = 20/27 ≈ 0.7407.

Pro tip: Emphasize that the green balls being returned means they are effectively 'delays' and do not affect the composition of the non-green balls. This simplifies the problem to a binomial with constant probability. Also, for the follow-up, note that the probability changes because the ratio of red to blue changes, not because of the total number of balls.

1. Clarify the process

Restate the problem: draw balls one at a time; green balls are returned and not kept; red and blue are kept. Stop when you have 3 kept balls. Ask if the process is clear.

2. Identify the distribution

Recognize that the green draws are irrelevant to the final composition; each kept ball is independently red with probability equal to the proportion of red among non-green balls. Thus the number of reds in 3 kept balls follows a Binomial distribution.

3. Compute for initial case

With 5 red and 5 blue, probability of red on a kept draw is 0.5. Compute P(X >= 2) = C(3,2)*(0.5)^3 + C(3,3)*(0.5)^3 = 0.5.

4. Compute for follow-up

With 10 red and 5 blue, probability of red is 10/15 = 2/3. Compute P(X >= 2) = C(3,2)*(2/3)^2*(1/3) + (2/3)^3 = 20/27 ≈ 0.7407.

5. Discuss implications

Explain that increasing the number of red balls increases the probability of getting at least 2 reds, and that the green balls being returned means they do not affect the relative proportions of red and blue.

Key Points to Mention

Green balls are returned, so they do not count toward the 3 held balls and do not change the composition of red and blue balls.
The number of red balls among the 3 kept balls follows a Binomial distribution with n=3 and p = (number of red)/(number of red + number of blue).
For the initial case, p = 5/10 = 0.5, so P(at least 2 red) = 0.5.
For the follow-up, p = 10/15 = 2/3, so P(at least 2 red) = 20/27 ≈ 0.7407.
The green balls being returned means they are effectively 'delays' and do not affect the probability of red on each kept draw.
The change in probability is due to the change in the ratio of red to blue, not the total number of balls.

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