Probability Theory

Probability Theory

Probability measures the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Business decisions inherently involve uncertainty — probability provides the mathematical framework for quantifying and managing that uncertainty.

Basic Concepts

An experiment is any process producing outcomes (tossing a coin, launching a product). The sample space (S) is the set of all possible outcomes. An event is a subset of the sample space. Mutually exclusive events cannot occur simultaneously (heads and tails on one toss). Exhaustive events cover all possibilities. Independent events: occurrence of one doesn’t affect the other (successive coin tosses). Complementary event: P(A’) = 1 − P(A).

Approaches to Probability

Classical (a priori): P(A) = favourable outcomes / total equally likely outcomes. Works for dice, cards, coins. Relative frequency (empirical): P(A) = number of times A occurred / total trials. Based on past data — used for insurance, quality control. Subjective: personal judgement based on experience — used when no historical data exists (probability of a new product succeeding). Axiomatic: formal mathematical framework defined by Kolmogorov’s axioms: P(A) ≥ 0, P(S) = 1, and P(A∪B) = P(A) + P(B) for mutually exclusive events.

Addition Rule

For mutually exclusive events: P(A or B) = P(A) + P(B). For non-mutually exclusive events: P(A or B) = P(A) + P(B) − P(A and B). The subtraction avoids double-counting the overlap. Example: P(drawing a king or a heart from a deck) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13. For three events: P(A∪B∪C) = P(A) + P(B) + P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C).

Multiplication Rule

For independent events: P(A and B) = P(A) × P(B). For dependent events: P(A and B) = P(A) × P(B|A) where P(B|A) is conditional probability. Example: probability of drawing two aces in succession without replacement = (4/52) × (3/51) = 12/2652 = 1/221. Independence means P(B|A) = P(B).

Conditional Probability

Conditional probability P(B|A) is the probability of B given that A has occurred: P(B|A) = P(A∩B)/P(A). Example: if 60% of customers are female and 25% are female professionals, then P(professional|female) = 0.25/0.60 = 0.417. Crucial in business: probability of loan default given recession, purchase probability given a website visit, defect probability given a specific supplier.

Bayes’ Theorem

Bayes’ theorem updates probabilities based on new evidence: P(A|B) = P(B|A) × P(A) / P(B). More generally for multiple hypotheses: P(Aᵢ|B) = P(B|Aᵢ)P(Aᵢ) / ΣP(B|Aⱼ)P(Aⱼ). Applications: updating machine fault probability based on defective output, revising market forecasts with new data, spam email filtering, medical diagnosis.

Counting Techniques

Multiplication principle: task 1 has m ways, task 2 has n ways → total m×n ways. Permutations (order matters): P(n,r) = n!/(n−r)!. Combinations (order doesn’t matter): C(n,r) = n!/[r!(n−r)!]. Example: choosing a 3-person committee from 10 = C(10,3) = 120 ways.

Summary

Probability theory — basic rules, addition and multiplication laws, conditional probability, Bayes’ theorem, and counting techniques — provides the framework for quantifying uncertainty, underpinning all inferential statistics, risk analysis, and decision-making under uncertainty.

Worked Example: Addition Rule

In a BBS class of 60 students, 35 study finance, 25 study marketing, and 10 study both. What is the probability that a randomly selected student studies finance or marketing?

Solution: P(F) = 35/60, P(M) = 25/60, P(F∩M) = 10/60

P(F or M) = P(F) + P(M) − P(F∩M) = 35/60 + 25/60 − 10/60 = 50/60 = 5/6 = 0.833

There is an 83.3% chance the student studies at least one of these subjects. Note: if we hadn’t subtracted the overlap (10 students studying both), we would have counted them twice, getting an incorrect answer of 60/60 = 1.

Worked Example: Bayes’ Theorem

A company buys components from three suppliers: Supplier A (50% of supply, 2% defective), Supplier B (30%, 3% defective), Supplier C (20%, 5% defective). If a randomly chosen component is defective, what is the probability it came from Supplier C?

Solution: Let D = defective event.

P(C|D) = P(D|C) × P(C) / P(D) = 0.010 / 0.029 = 0.345 or 34.5%

Interpretation: Although Supplier C provides only 20% of components, it is responsible for 34.5% of defective items because of its higher defect rate. The company should work with Supplier C to improve quality or consider changing suppliers. This is the power of Bayes’ theorem — it updates our initial assessment (20% from C) with new evidence (the item is defective) to get a revised probability (34.5%).

Exam Tips for Probability

Tip 1: Draw a Venn diagram for addition rule problems — it prevents double-counting errors. Tip 2: For Bayes’ theorem, always set up a table with prior probabilities and likelihoods — it organises the calculation clearly. Tip 3: Check your answer: probabilities must be between 0 and 1, and P(A) + P(A’) must equal 1. Tip 4: "At least one" problems are easiest solved as 1 − P(none). Tip 5: For independent events, P(A∩B) = P(A)×P(B). If this doesn’t hold, events are dependent — use conditional probability.