Seven white balls and three black balls are placed randomly in a row. What is the probability that no two black balls are adjacent?

Introduction / Context:
This is a classic “no two alike adjacent” arrangement probability. Treat same-colored balls as indistinguishable, and count ways to position the black balls among the whites so that adjacency is avoided.

Given Data / Assumptions:

• 7 white (W) balls, 3 black (B) balls.
• Balls of the same color are indistinguishable for counting positions.
• Random linear arrangement among 10 positions.

Concept / Approach:
First arrange the 7 W’s. They create 8 “gaps” (including ends). To avoid adjacency, place the 3 B’s in 3 distinct gaps. Total ways to choose 3 gaps from 8 is C(8,3). Total ways to place 3 B’s among 10 slots (without restriction) is choosing 3 positions for B out of 10: C(10,3). The desired probability is C(8,3) / C(10,3).

Step-by-Step Solution:

Verification / Alternative check:
Direct small enumeration for a reduced case (e.g., 3 W and 2 B) gives identical “gap” logic, supporting the method. The gap method is a standard combinatorial approach for adjacency constraints.

Why Other Options Are Wrong:
1/2, 1/3, and 2/15 do not match the ratio of admissible to total placements; they stem from counting errors or treating balls as distinct when they are not.

Common Pitfalls:
Forgetting that black balls must occupy distinct gaps (placing two in the same gap creates adjacency), or incorrectly counting permutations of indistinguishable balls.

Final Answer:
7/15