A bag has 5 blue and 4 black balls. Three balls are drawn at random without replacement. What is the probability that exactly two are blue and one is black?

Introduction / Context:This is a standard hypergeometric probability: sampling without replacement from a finite population containing two types (blue and black). We compute the probability of drawing a specified composition.

Given Data / Assumptions:

• Total balls = 9 (5 blue, 4 black).
• Draw 3 balls without replacement.
• Desired outcome: exactly 2 blue and 1 black.

Concept / Approach:Hypergeometric probability: P = [C(blue, needed_blue) * C(black, needed_black)] / C(total, draws). Here needed_blue = 2 and needed_black = 1. No order factor is needed because combinations already account for unordered draws.

Step-by-Step Solution:

Verification / Alternative check:Decimal check: 10/21 ≈ 0.4762. A quick simulation mindset (blue fraction ≈ 0.556) suggests the most likely triple roughly contains 2 blues, supporting plausibility of ≈0.48.

Why Other Options Are Wrong:2/5 (=0.4) and 1/3 (=0.333…) underestimate because two-blue outcomes are relatively frequent given more blues than blacks; 1/6 is far too small.

Common Pitfalls:Accidentally multiplying by permutations (overcounting), or using replacement formulas; also mixing numerator terms (e.g., C(4,2) instead of C(5,2)).

Final Answer:10/21