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

Introduction / Context:
With indistinguishable balls of two colors, the number of distinct linear arrangements equals the number of ways to choose positions for the black balls among 10 slots. We require that no two black balls be adjacent.

Given Data / Assumptions:

• Whites W = 7 (identical), blacks B = 3 (identical).
• Total arrangements (distinguishing only by color) = C(10, 3) by choosing positions for B's among 10 slots.
• Constraint: B's are non-adjacent.

Concept / Approach:
Place 7 W first, creating 8 gaps (including ends). To avoid adjacency, place each B in a distinct gap. Choose 3 of the 8 gaps to host one B each.

Step-by-Step Solution:
Favorable = C(8, 3) (choose which gaps receive a B).Total = C(10, 3) (choose any 3 positions for B among 10).Probability = C(8, 3) / C(10, 3) = 56 / 120 = 7/15.

Verification / Alternative check:
Compute combinations: C(8, 3) = 56; C(10, 3) = 120; reduce the fraction.

Why Other Options Are Wrong:
1/2 and 1/3 are rough guesses; 2/15 undercounts; 1/5 ignores the combinatorial structure.

Common Pitfalls:
Forgetting to include end gaps or attempting to “permute” identical balls.

Final Answer:
7/15