Day 1 · Dot Placement¶
Edited by Charles
| Study record | Value |
|---|---|
| Question | 4 of 4 |
| Topic | Probability |
| Difficulty | Easy |
| Attempts | 1 |
| Mastery (Notion notation) | ❚ |
| Latest attempt | October 22, 2026 |
| Source | OpenQuant: Dot Placement |
Probability of Three Dots on Distinct Edges¶
Problem¶
You place three dots independently and uniformly along the edges of a regular octagon. What is the probability that all three dots lie on distinct edges?
Solution: Sequential Probability¶
A regular octagon has 8 edges, each equally likely to contain a dot.
- First dot: Any edge → P₁ = 8/8 = 1
- Second dot: Must choose a different edge → P₂ = 7/8
- Third dot: Must avoid the previous two edges → P₃ = 6/8
Multiplying the conditional probabilities:
P = (8/8) × (7/8) × (6/8) = 21/32 = 65.625%
Alternative: Combinatorics¶
- Total outcomes: 8³
- Favorable outcomes: 8 × 7 × 6
P = (8 × 7 × 6)/8³ = 21/32
General Formula¶
For n equally likely edges and k independently placed dots (k ≤ n):
P(all distinct) = n! / [(n − k)! × nᵏ]
Key Takeaways¶
- Technique: Conditional Probability / Counting
- Concept: Sampling with Replacement, Birthday Problem
- Interview Tip: Use sequential conditional probabilities for the quickest solution.
- Assumption: All edges have equal probability, and dot placements are independent.