AIML Rounds
Interview20 minBefore this:Probability Basics & Bayes' Rule

Balls & Urns Practice

With or without replacement, and the symmetry shortcut that saves minutes.

How to use this page

Give yourself about 5 minutes per question, as in an interview. Say your approach out loud, then open the answer and compare the method, not just the number.
The one tool for this page: Without replacement, multiply step by step (or use combinations); with replacement, draws are independent. Symmetry often beats calculation: every ball is equally likely to be in any position.

Question 1 · Easy · without replacement

An urn has 5 red and 3 blue balls. You draw 2 without replacement. What is the probability both are red? And with replacement?

Say your approach out loud first.

Answer: 5/14 ≈ 0.357 without; 25/64 ≈ 0.391 with

  1. Without: 5/8 × 4/7 = 20/56 = 5/14.
  2. With: 5/8 × 5/8 = 25/64.
  3. Without replacement is lower because the first red removes a red.

Question 2 · Easy · last ball

A bag has 3 red and 2 green balls. You draw them all one by one. What is the probability the last ball is green?

Say your approach out loud first.

Answer: 2/5

  1. By symmetry, every ball is equally likely to be last.
  2. 2 of the 5 balls are green: 2/5. The same as the first ball being green.

Question 3 · Medium · which urn

Urn A has 2 red and 1 blue ball; urn B has 1 red and 3 blue. You pick an urn at random and draw a red ball. What is the probability it came from urn A?

Say your approach out loud first.

Answer: 8/11 ≈ 0.73

  1. P(red | A) = 2/3, P(red | B) = 1/4, each urn chosen with ½.
  2. P(A | red) = (½ · 2/3) / (½ · 2/3 + ½ · 1/4) = (1/3) / (11/24) = 8/11.

Question 4 · Medium · Pólya urn

An urn starts with 1 red and 1 blue ball. Each time you draw a ball, you put it back along with another ball of the same colour. What is the probability that the first two draws are both red?

Say your approach out loud first.

Answer: 1/3

  1. First draw red: 1/2. Then the urn has 2 red, 1 blue, so second red: 2/3.
  2. 1/2 × 2/3 = 1/3.
  3. Surprising fact: after n draws, the number of reds drawn is uniformly distributed on 0…n.

Question 5 · Medium · empty boxes

You throw 10 balls independently into 10 boxes, each box equally likely. How many boxes do you expect to stay empty?

Say your approach out loud first.

Answer: 10 × 0.9¹⁰ ≈ 3.49

  1. A given box stays empty if all 10 balls miss it: (9/10)¹⁰ ≈ 0.349.
  2. By linearity of expectation over the 10 boxes: 10 × 0.349 ≈ 3.49.
  3. For large n, about n/e boxes stay empty (≈ 36.8%).

Follow-up: This is the same maths as the share of rows a bootstrap sample leaves out (out-of-bag rows in a Random Forest): about 36.8%.

Question 6 · Hard · same colour pair

An urn has 4 red and 4 blue balls. You draw 4 without replacement. What is the probability you get exactly 2 of each colour?

Say your approach out loud first.

Answer: 18/35 ≈ 0.51

  1. Ways to choose 2 red and 2 blue: C(4,2) × C(4,2) = 36.
  2. All ways to choose 4 of 8: C(8,4) = 70.
  3. 36/70 = 18/35 (the hypergeometric distribution).
More practice and sources

6 interview questions, easy to hard · answers hidden until you click · checked 28 Oct 2026

Keep practising

These are classic interview questions. For more in the same style: Frederick Mosteller, Fifty Challenging Problems in Probability with Solutions (Dover, 1987), and Xinfeng Zhou, A Practical Guide to Quantitative Finance Interviews (2008). Related lessons: Cross-Validation (sampling).