Question: A mechanical engineer is testing 9 parts, 4 of which are from Supplier A and 5 from Supplier B. If she randomly selects 3 parts, what is the probability that all 3 are from different suppliers? - Treasure Valley Movers

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**Understanding Probability in Real-World Supplier Selection One of the quieter but increasingly discussed challenges in manufacturing and quality control is probabilistic compatibility in supply chains—specifically, when a mechanical engineer tests multiple components from mixed sources. A common scenario involves evaluating 9 parts: 4 from Supplier A and 5 from Supplier B. If selected randomly, what’s the chance all three tested parts come from different suppliers? This question reveals practical applications of probability that matter to engineers, quality analysts, and procurement teams shaping production reliability. Though seemingly simple, exploring this helps clarify decision-making in risk assessment and supplier diversification.

Why This Question Matters in US Industries Today
With supply chains under constant pressure from geopolitical shifts, material costs, and quality consistency, understanding component sourcing reliability is critical. In engineering fields—from automotive to aerospace—teams rely on predictable performance, yet supplier variability remains a blind spot for many. This probability problem reflects real-world testing workflows where engineers rigorously validate part performance across multiple sources, seeking to identify imbalance or overdependence. The query highlights a growing awareness of data-driven validation: ensuring no single source dominates testing, which could skew reliability benchmarks and invite quality risks.

How to Calculate the Probability: Breaking It Down
To determine the likelihood all three randomly selected parts come from different suppliers, we use combinatorics—mapping possible selections and identifying favorable outcomes. The engineer picks 3 parts from a total of 9 without replacement. The total number of ways to choose 3 parts from 9 is given by 9 choose 3:
9! / (3! × (9–3)!) = 84 unique combinations.
For all parts to be from different suppliers, one must come from Supplier A and two from Supplier B—or vice versa. Since there are only two suppliers, full diversity across three parts demands choosing exactly one from A and two from B, or two from A and one from B—depending on which supplier contributes two selections. However, with only 4 parts from A and 5 from B, selecting two from Supplier A is feasible, while selecting two or more from Supplier B—up to 3—remains possible due to sufficient volume.

Favorable outcomes require:

• Exactly 1 from Supplier A and 2 from Supplier B
• Or exactly 2 from Supplier A and 1 from Supplier B

Calculate each:

• Choosing 1 from A: 4C1 = 4

• Choosing 2 from B: 5C2 = 10
  → 4 × 10 = 40 ways for 1A + 2B

• Choosing