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BUSINESS STATISTICSβ€ΊUNIT 03β€ΊPROBABILITY

Unit 03 β€” Theories of Probability

Classical theory, relative frequency theory, axioms, addition rule, multiplication rule, rule of at least one, concept of expected number of successes, Bayes' theorem.

SESSIONS: 2 TEXTBOOK: LEVIN CH. 4, pp. 153–206 EXAM WEIGHT: Q5 (a) β€” 5 MARKS, IN BOTH PAPERS Full PPT: Unit 3 deck 44 Bayes notes 3
PRACTISE THIS UNIT

Every question shape in this unit is drilled on Unit 03 β€” How to Solve Anything: four topics, each with a "how to recognise this topic" card, a master method, and every exercise Levin sets solved with its page number. Topic 4 is Q5 (a) β€” Bayes, 5 marks, worked as a five-line table.

β—† What the papers ask from this unit

One question, 5 marks, in both papers, and it is always the same three-part shape: a classifier (credit-risk model in the Final, churn model in the Re-Exam) with a base rate and two accuracy percentages, then "the model flagged someone β€” what is the probability they really are what it says?", then "what is the managerial insight?".

That is Bayes' theorem with a mandatory sentence of business English at the end. The insight sentence carries marks β€” both official keys write one out in bold. Final Q5 (a) QP p. 2, Re-Exam Q5 (a) QP p. 2. Both solved β†’

1. The Three Theories of Probability Core syllabus concept

1Understand the Concept

Probability is "a quantitative description of the chances or likelihoods associated with various outcomes" and it "provides a bridge between descriptive and inferential statistics" slide 1. On the scale from 0 to 1: 0 = the event is very unlikely, 0.5 = as likely as not, 1 = almost certain slide 3.

The syllabus names the theories; the deck calls them "methods of assigning probabilities" slide 11. Same three things:

TheoryBasisAssignsDeck's example
Classical The assumption that all outcomes are equally likely. Probability 1/n to each of n outcomes. Rolling a die: S = {1,2,3,4,5,6}, each outcome has a 1/6 chance slide 12.
Relative frequency Experimentation or historical data. frequency Γ· total frequency. Lucas Tool Rental: 18 of the last 40 days saw 2 polishers rented β†’ P(2) = 18/40 = .45 slide 14.
Subjective Judgement. Whatever the analyst believes, provided the axioms hold. An analyst's probabilities over the eight Bradley Investments outcomes slide 15.

2The vocabulary the exam assumes you have

  • An experiment is any process that generates well-defined outcomes.
  • The sample space is the set of all experimental outcomes.
  • An experimental outcome is also called a sample point slide 4.
  • An event is a collection of sample points, and its probability is the sum of the probabilities of the sample points in it slide 16.
  • A simple event is determined by the outcome of one trial; a compound event is a combination of at least two simple events slide 17.

3The four probability types β€” a favourite short-answer question

TypeDefinition slide 31
A prioriA probability estimate made prior to receiving new information.
PosteriorA probability that has been revised after additional information was obtained. ← this is what Bayes produces
MarginalThe unconditional probability of one event occurring; the probability of a single event.
JointThe probability of two events occurring together or in succession.

4Counting rules β€” when the sample space is too big to list

Multi-step Experiment
\[ \text{Total outcomes} = n_1 \cdot n_2 \dots n_k \]
Combinations (Order doesn't matter)
\[ C(N,n) = \binom{N}{n} = \frac{N!}{n!(N-n)!} \]
Permutations (Order matters)
\[ P(N,n) = n! \cdot C(N,n) = \frac{N!}{(N-n)!} \qquad (N! = N(N-1)\dots 1, \; 0! = 1) \]

slides 9–10. The combination formula reappears immediately in Unit 4 as the nCr in the binomial formula.

ENMultiplication & addition rule β€” mutually exclusive & independent events The Organic Chemistry Tutor
Sources
Primary Unit 3 deck β€” likelihood scale, experiments, counting rules, three methods slides 1–18
Textbook Levin Β§4.2–4.3 Basic terminology; three types of probability pp. 155–163

2. The Addition and Multiplication Rules Core syllabus concept

1Set language first β€” it decides which rule you need

TermSymbolEnglish triggerMeaning
ComplementAc"not A", "none", "fails"All sample points not in A slide 22.
UnionA βˆͺ B"or", "either", "at least one of"All points in A, or B, or both slide 23.
IntersectionA ∩ B"and", "both"All points in both A and B slide 25.
Mutually exclusiveA ∩ B = βˆ…"cannot both happen"No sample points in common; if one occurs the other cannot slide 29.

2The rules

Addition Law ("or")
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \qquad \text{(Mutually exclusive: } P(A \cup B) = P(A) + P(B)\text{)} \]
Multiplication Law ("and")
\[ P(A \cap B) = P(B) \cdot P(A \mid B) \qquad \text{(Independent: } P(A \cap B) = P(A) \cdot P(B)\text{)} \]
Conditional Probability
\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]
Complement
\[ P(A^c) = 1 - P(A) \]

Addition slide 27, mutually exclusive form slide 30, conditional slide 32, multiplication slide 34, independent form slide 37.

⚠ Mutually exclusive β‰  independent

These are opposites, not synonyms. Mutually exclusive means knowing A happened tells you B definitely did not β€” that is maximum dependence. Independent means knowing A happened tells you nothing about B. Two events with non-zero probabilities can never be both.

3Worked example β€” Bradley Investments, all four rules on one dataset

M = Markley Oil profitable, C = Collins Mining profitable, with the analyst's subjective probabilities from slide 15.

  1. Step 1 β€” Marginal probabilities, by summing sample points P(M) = .20 + .08 + .16 + .26 = 0.70  β† slide 19 P(C) = .20 + .16 + .10 + .02 = 0.48  β† slide 20
  2. Step 2 β€” Joint probability, by intersection M ∩ C = {(10,8), (5,8)} β†’ P(M ∩ C) = .20 + .16 = 0.36  β† slide 26
  3. Step 3 β€” Union, two ways, to prove the addition law works By listing: P(M βˆͺ C) = .20+.08+.16+.26+.10+.02 = 0.82 By the law: P(M βˆͺ C) = 0.70 + 0.48 βˆ’ 0.36 = 0.82 βœ“  β† slide 28
  4. Step 4 β€” Conditional probability P(C | M) = P(C ∩ M) / P(M) = 0.36 / 0.70 = 0.5143  β† slide 33

    Read it in English: given that Markley Oil is profitable, there is a 51.43% chance Collins Mining is too.

  5. Step 5 β€” Test for independence Does P(M ∩ C) = P(M)Β·P(C)? P(M)Β·P(C) = 0.70 Γ— 0.48 = 0.336 β‰ˆ 0.34, but P(M ∩ C) = 0.36

    0.34 β‰  0.36, so M and C are NOT independent slide 38. This is the standard way an exam asks you to test independence β€” compute both sides and compare.

ENBasic probability: the multiplication rule jbstatistics ENAddition and multiplication rules of probability DrCraigMcBridePhD

Practice C1–C4 β€” addition, multiplication, conditional, independence β†’

Sources
Primary Unit 3 deck β€” complement, union, intersection, addition, conditional, multiplication, independence slides 21–38
Textbook Levin Β§4.4–4.6 Probability rules; independence; dependence pp. 164–188

3. The Rule of At Least One Core syllabus concept

1Understand the Concept

This is named separately in the syllabus Syllabus p. 1 because it has its own trigger phrase and its own trick. Computing "at least one" directly means adding the probabilities of exactly one, exactly two, exactly three… Computing its complement means one multiplication.

Your class notes write the rule in red, and it is worth copying exactly Class notes p. 9:

"If AT LEAST β†’ always 1 βˆ’ P(getting none)."
Rule of At Least One
\[ P(\text{at least 1 success}) = 1 - P(\text{no successes}) = 1 - (1-p_1)(1-p_2)\dots(1-p_n) \]
All p equal
\[ P(\text{at least 1}) = 1 - (1-p)^n \]

2Worked example β€” the job-applications case, solved in class

This is Case 2 from your case pack, and your class worked it Class notes p. 9. A business graduate applies to the top 10 accounting firms. Applying is costly. The cost and the probability of an offer at each firm are tabulated in decreasing order of cost.

Company12345678910
Cost ($)870600540500400320300230200170
P(offer).38.35.28.20.18.18.17.14.14.08
  1. Q1 β€” If she applies to all 10, what is P(at least one offer)?

    Do not add the probabilities β€” they would sum to 2.10, which is impossible. Take the complement:

    P(at least one) = 1 βˆ’ P(no offer anywhere) = 1 βˆ’ [(1βˆ’.38)(1βˆ’.35)(1βˆ’.28)(1βˆ’.20)(1βˆ’.18)(1βˆ’.18)(1βˆ’.17)(1βˆ’.14)(1βˆ’.14)(1βˆ’.08)] = 1 βˆ’ [0.62 Γ— 0.65 Γ— 0.72 Γ— 0.80 Γ— 0.82 Γ— 0.82 Γ— 0.83 Γ— 0.86 Γ— 0.86 Γ— 0.92] = 1 βˆ’ 0.0881 = 0.912

    About a 91% chance of at least one offer, at a total cost of $3,930. Your class notes record 0.914 for this β€” the small gap is rounding, since they kept fewer decimals through the ten-way product. Either figure is fine; show the working.

  2. Q3 β€” If she applies to companies 2, 5, 8 and 9 only Total cost = 600 + 400 + 230 + 200 = $1,430 P(at least one) = 1 βˆ’ (0.65)(0.82)(0.86)(0.86) = 1 βˆ’ 0.394 = 0.606

    Your class notes record 0.6057 for this style of sub-question Class notes p. 9.

  3. Q4 β€” Reach 75% confidence at minimum cost

    You need 1 βˆ’ Ξ (1βˆ’pα΅’) β‰₯ 0.75, i.e. Ξ (1βˆ’pα΅’) ≀ 0.25. To make that product small for the least money, add companies in order of "probability bought per dollar", not raw probability. Company 10 gives .08 for $170; company 1 gives .38 for $870. Work down the value ranking, accumulate the product of (1βˆ’pα΅’), and stop the moment it drops below 0.25.

  4. Q5 β€” Maximise the chance on a $1,500 budget

    Same logic, opposite constraint: choose the subset whose costs total ≀ $1,500 and whose product Ξ (1βˆ’pα΅’) is smallest. This is a knapsack-style search over a handful of options, done by trial.

3Expected number of successes

Also named in the syllabus. It answers "how many, on average" rather than "what are the chances":

Expected Number of Successes
\[ E(X) = \sum p_i \qquad \text{or} \qquad E(X) = n \cdot p \]

For all 10 applications: E(offers) = 0.38 + 0.35 + … + 0.08 = 2.10 offers. Notice this is a perfectly sensible answer even though it exceeds 1 β€” an expected count is not a probability. The nΒ·p form is the mean of the binomial distribution, which is where Unit 4 picks it up.

⚠ Common mistakes
  • Adding probabilities for "at least one". That only works when the events are mutually exclusive, and separate job applications are not.
  • Forgetting to subtract from 1. The product (1βˆ’p₁)(1βˆ’pβ‚‚)… is P(none). You are one step from the answer, not at it.
  • Confusing the expected number with the probability. 2.10 expected offers and a 0.914 probability of at least one offer are both correct and describe the same situation.

Practice C5–C6 β€” at least one, and expected number of successes β†’

Sources
Primary Class notes β€” "If at least β†’ always 1 βˆ’ P(getting none)", case worked p. 9
Case Case 2 β€” accounting-firm applications, from the probability case pack
Textbook Levin Β§4.5 Probabilities under statistical independence pp. 170–178

4. Law of Total Probability and Bayes' Theorem Core syllabus concept Β· highest exam value

1Understand the Concept

Bayes' theorem is a machine for reversing a conditional probability. You are told P(evidence | cause) and you want P(cause | evidence). Every exam version of this question is the same reversal: the model tells you P(flagged | actually high-risk), and you are asked for P(actually high-risk | flagged).

It comes in two steps, and your class notes write both out in full Class notes p. 1.

Step 1 β€” Total Probability
\[ P(A) = \sum_{j=1}^k P(S_j) \cdot P(A \mid S_j) = P(S_1)P(A \mid S_1) + \dots + P(S_k)P(A \mid S_k) \]
Step 2 β€” Bayes' Theorem
\[ P(S_i \mid A) = \frac{P(S_i) \cdot P(A \mid S_i)}{\sum P(S_j) \cdot P(A \mid S_j)} = \frac{P(S_i) \cdot P(A \mid S_i)}{P(A)} \]
The proof, in two lines Class notes pp. 1–2: from the definition of conditional probability, P(A|Sα΅’) = P(A∩Sα΅’)/P(Sα΅’), so P(A∩Sα΅’) = P(Sα΅’)Β·P(A|Sα΅’). Substituting that into P(Sα΅’|A) = P(Sᡒ∩A)/P(A), and replacing P(A) with the total-probability sum, gives the theorem. If a question says "state and prove Bayes' theorem", that is the whole answer.

2Method β€” the five lines that always work

5-Step Bayes Routine
\[ 1. \text{Define events} \quad 2. \text{Write priors \& conditionals} \quad 3. P(A) \text{ Total Prob} \quad 4. P(S_i \mid A) \text{ Bayes} \quad 5. \text{Managerial Insight} \]
⚠ Reading the false-positive rate

The two papers phrase the same number in opposite ways, and getting this wrong wrecks the denominator:

  • Final: "correctly classifies 90% of Low-Risk applicants as Low Risk" β†’ the model wrongly flags 10% of them, so P(T|L) = 1 βˆ’ 0.90 = 0.10.
  • Re-Exam: "incorrectly flags 20% of loyal customers as churners" β†’ that is the false-positive rate already, so P(P|L) = 0.20. No subtraction.

3Worked example β€” the deck's heart-attack problem

Your class solved this one too Class notes p. 2. In a region, 49% of the population are female. Of the female patients, 8% are high risk for heart attack; 12% of the male patients are high risk. A person is selected at random and found to be high risk. What is the probability it is a male?

  1. Step 1 β€” Define F = selected person is female   M = selected person is male   H = high risk of heart attack
  2. Step 2 β€” Priors and conditionals P(F) = 0.49    P(M) = 0.51    P(H|F) = 0.08    P(H|M) = 0.12
  3. Step 3 β€” Total probability P(H) = P(M)P(H|M) + P(F)P(H|F) = (0.51)(0.12) + (0.49)(0.08) = 0.0612 + 0.0392 = 0.1004
  4. Step 4 β€” Bayes P(M|H) = (0.51)(0.12) / 0.1004 = 0.0612 / 0.1004 = 0.61 = 60%
  5. Step 5 β€” Interpret

    Males are 51% of the population but 61% of the high-risk group β€” the prior of 51% has been revised upward by the evidence. That gap between prior and posterior is the entire content of a Bayes answer.

4The rare-disease problem β€” why the answer feels wrong

The deck closes with the example that makes Bayes memorable slides 43–44. A disease infects 1 in 1000. The test catches 99% of infected people, but also returns a positive for 2% of healthy people. Someone tests positive. What are their chances of having the disease?

  1. Given P(A) = 0.001   P(Ac) = 0.999   P(B|A) = 0.99   P(B|Ac) = 0.02
  2. Apply P(A|B) = (0.001)(0.99) / [(0.001)(0.99) + (0.999)(0.02)] = 0.00099 / (0.00099 + 0.01998) = 0.00099 / 0.02097 = 0.0472
  3. Interpret

    A "99% accurate" test returns a positive that is right only 4.7% of the time. The reason: the disease is so rare that the 2% of false positives drawn from the huge healthy group (about 20 people per 1000) swamp the 1 true positive. When the base rate is tiny, most positives are false positives.

    That single sentence is the managerial insight both papers ask for, restated for a credit model or a churn model.

Insight sentences the official keys accepted
Final: "Even though the model classifies an applicant as high risk, 1 in 3 such classifications are false alarms, suggesting the bank should review borderline cases manually before rejecting them." Key p. 4
Re-Exam: "Only about one-third of the flagged customers truly churn. Thus, targeted retention offers should focus on this smaller segment to optimize marketing spend instead of giving discounts to all flagged users." Key p. 4
Both follow the same template: quantify the false-alarm rate β†’ name one concrete action.
ENBayes' Theorem, clearly explained StatQuest with Josh Starmer ENBayes theorem made easy (Business statistics) Unscripted life ENBayes theorem with a worked example Simplilearn

Practice D1–D6 β€” six Bayes problems in the exam's exact shape β†’

Sources
Primary Class notes β€” total probability, Bayes, the entire proof, heart-attack example pp. 1–2
Lecture Unit 3 deck β€” total probability, Bayes' rule, both examples slides 39–44
Textbook Levin Β§4.7 Revising prior estimates of probabilities: Bayes' theorem pp. 188–196
Exam Final Q5 (a) p. 2 Β· Re-Exam Q5 (a) p. 2