Basic Probability Concepts

Basic Probability Concepts (CSSGB BoK III.B.1 — Understand)

Quick Answer: Probability quantifies how likely an event is. Green Belts need independence vs. mutual exclusivity, addition and multiplication rules, and permutations vs. combinations so later work on distributions, sampling, DPMO logic, and hypothesis testing rests on solid foundations. Understand-level means you can interpret and compute standard problems—not derive advanced theory.

Probability Language

An experiment is a repeatable process with uncertain outcomes (inspect a unit, roll a die, sample a lot). The sample space S is the set of all possible outcomes. An event A is a subset of S.

• P(A) satisfies 0 ≤ P(A) ≤ 1 and P(S) = 1.
• Complement: A′ (not A) has P(A′) = 1 − P(A).
• Equally likely outcomes: if S has n outcomes and A has k of them, P(A) = k/n (classical model).

In quality work, probabilities often come from relative frequencies (historical defect rates) or models (binomial, normal) rather than dice. The algebra is the same.

Mutually Exclusive Events

Events A and B are mutually exclusive (disjoint) if they cannot occur on the same trial: A ∩ B = ∅, so P(A and B) = 0.

Addition rule (mutually exclusive):

P(A or B) = P(A) + P(B)

General addition rule (any events):

P(A or B) = P(A) + P(B) − P(A and B)

The subtraction removes double-counting outcomes that sit in both events.

Worked example — exclusive defect codes: On one unit, inspection assigns at most one primary defect code: scratch (probability 0.04), dent (0.03), or wrong label (0.02). These primary codes are treated as mutually exclusive.

• P(scratch or dent) = 0.04 + 0.03 = 0.07
• P(any of the three) = 0.04 + 0.03 + 0.02 = 0.09
• P(no primary defect among these) = 1 − 0.09 = 0.91

If a unit could have both a scratch and a wrong label counted as two events of interest, they are not mutually exclusive and you must use the general addition rule with a known P(both).

Independent Events

Events A and B are independent if knowing that A occurred does not change the probability of B:

P(B|A) = P(B) (and equivalently P(A|B) = P(A))

Multiplication rule for independent events:

P(A and B) = P(A) × P(B)

Critical exam trap: Independent ≠ mutually exclusive. If two events with positive probability are mutually exclusive, then P(A and B) = 0, which cannot equal P(A)P(B) > 0. So non-trivial mutually exclusive events are dependent.

Worked example — independent stations: Station A fails a check with probability 0.02; Station B fails with probability 0.03. Failures are independent.

• P(both fail) = 0.02 × 0.03 = 0.0006
• P(A fails and B passes) = 0.02 × (1 − 0.03) = 0.02 × 0.97 = 0.0194
• P(at least one fails) = 1 − P(both pass) = 1 − (0.98 × 0.97) = 1 − 0.9506 = 0.0494

Using complements for “at least one” is often faster than expanding all failure combinations.

Conditional Probability and the General Multiplication Rule

Conditional probability:

P(B|A) = P(A and B) / P(A) when P(A) > 0

General multiplication rule:

P(A and B) = P(A) × P(B|A) = P(B) × P(A|B)

Worked example — dependent lots: A bin has 10 parts: 3 nonconforming, 7 conforming. Draw two parts without replacement.

• P(first nonconforming) = 3/10 = 0.3
• P(second nonconforming | first nonconforming) = 2/9 ≈ 0.222
• P(both nonconforming) = (3/10) × (2/9) = 6/90 = 0.0667

If sampling were with replacement (or a huge lot so depletion is negligible), the draws would be treated as independent and P(both nonconforming) ≈ 0.3 × 0.3 = 0.09—different answer, different assumption.

Permutations and Combinations

Counting rules feed probability numerators and denominators when outcomes are equally likely.

Factorial: n! = n × (n − 1) × … × 1, with 0! = 1.

Permutations — order matters; arrange r items from n distinct items:

nPr = n! / (n − r)!

Combinations — order does not matter; choose r items from n:

nCr = C(n, r) = n! / (r!(n − r)!)

Worked example — permutations: How many ways can a team assign first, second, and third place among 5 improvement ideas?

• 5P3 = 5! / (5 − 3)! = 120 / 2 = 60

Worked example — combinations: A lot has 12 cartons; an auditor randomly selects 3 for full inspection. How many possible samples?

• 12C3 = 12! / (3! × 9!) = (12 × 11 × 10) / (3 × 2 × 1) = 1320 / 6 = 220

Worked probability with combinations: Among 12 cartons, 2 are nonconforming. P(the sample of 3 contains both nonconforming cartons):

• Ways to choose both bad and 1 good: C(2,2) × C(10,1) = 1 × 10 = 10
• Total samples: C(12,3) = 220
• Probability = 10 / 220 = 1/22 ≈ 0.0455

Connecting Probability to Green Belt Practice

• Independent trials underwrite binomial models for defectives when p is constant and trials do not affect each other.
• “At least one defect” calculations appear in risk and multi-opportunity thinking (related to why multiple opportunities raise overall unit failure chance).
• Sampling without replacement appears in finite-lot inspection; large-lot approximations restore near-independence.
• Combinations explain why random sampling plans enumerate subsets, not sequences.

Common Mistakes to Avoid on the Exam

1. Adding probabilities for “and” instead of multiplying (or the reverse).
2. Treating mutually exclusive events as independent.
3. Using nPr when the problem only cares about which items were selected.
4. Forgetting to subtract the intersection in the general “or” rule.
5. Dropping the complement approach when “at least one” expands into many cases.

Master these five ideas—complement, exclusive addition, independent multiplication, conditional multiplication, and nPr vs. nCr—and you have the BoK III.B.1 toolkit for CSSGB Measure statistics.