BUSINESS STATISTICS›PROBLEM SOLVING›UNIT 03

Unit 03 — How to Solve Anything

Four topics that build in a straight line: what probability is, then OR, then AND, then reversing a conditional. That last one is Q5 (a) — 5 marks in both papers, and it is a five-line table you can learn by shape.

TOPICS: 4 TEXTBOOK: LEVIN CH. 4, pp. 153–199 EXAM WEIGHT: Q5 (a) BAYES — 5 MARKS PAIRS WITH: Unit 03 notes

◆ The whole unit on one line each

OR (addition) P(A or B) = P(A) + P(B) − P(AB) subtract the overlap mutually exclusive → P(AB) = 0, so just add AND (multiplication) independent: P(AB) = P(A) × P(B) dependent: P(AB) = P(A|B) × P(B) GIVEN (conditional) independent: P(B|A) = P(B) knowing A tells you nothing dependent: P(B|A) = P(AB) / P(A) REVERSING (Bayes) P(Event|Evidence) = P(Event) × P(Evidence|Event) ─────────────────────────── P(Evidence)

Every question in this unit is one of those four, or a chain of them. The skill is reading which — and Topic 4's ID card is the one that matters most, because Bayes questions disguise themselves as ordinary conditional-probability questions.

Topic 1 of 4 · Levin §4.1–§4.3 · pp. 155–164 · 14 textbook questions

Probability Basics — Sample Space and the Three Theories

The vocabulary layer. Two things get examined: counting outcomes to get a classical probability, and naming which of the three approaches produced a given number. Both are quick marks if you have the definitions ready.

How to know a question belongs to this topic

It asks you to

Count favourable outcomes over total outcomes, list a sample space, or classify where a probability came from.

Give-away words

drawing one card from a deck of 52 collectively exhaustive mutually exclusive sample space classify as classical, relative frequency or subjective based on these data, what is the probability

The three theories — how to tell them apart

Classical: derived from the structure of the experiment before any data — cards, dice, coins. Equally likely outcomes, counted.
Relative frequency: derived from observed data — "209 days functioning out of 260".
Subjective: derived from judgement — a one-off event that cannot be repeated or counted.

Don't confuse it with

If the question…	then…
asks for "A or B"	Topic 2 — addition rule
asks for "A and B"	Topic 3 — multiplication rule
says "given that…" and then reverses the order	Topic 4 — Bayes
just counts outcomes or names a method	Here

Master Method — Topic 1

Formula bank

Classical probability

\[ P(\text{event}) = \frac{\text{number of outcomes where the event occurs}}{\text{total number of possible outcomes}} \]

Relative frequency

\[ P(\text{event}) = \frac{\text{number of times it occurred}}{\text{total number of trials}} \]

Axioms

\[ 0 \le P(A) \le 1 \qquad \sum P(\text{all outcomes}) = 1 \qquad P(A') = 1-P(A) \]

Mutually exclusive = cannot both happen at once. Collectively exhaustive = between them they cover every possibility. A list can be one, both, or neither.

Traps

Confusing mutually exclusive with independent. They are almost opposites: mutually exclusive events are strongly dependent — if one happens, the other definitely did not.
Miscounting a deck. 52 cards, 4 suits of 13, 26 red / 26 black, 12 face cards (J, Q, K), 4 of each rank.
Forgetting that probabilities must sum to 1 over a collectively exhaustive, mutually exclusive list.

Textbook bank §4.1–§4.3, pp. 155–164 · solved

14 exercises Levin §4.1 p. 153 · §4.2 p. 156 · §4.3 p. 158

4-1TheoryLevin Ex. 4-1 · p. 155

Q. Life insurers know for certain that every policyholder is going to die. Does this mean probability theory does not apply to life insurance?

AnswerNo — probability theory is exactly what the business runs on. The insurer is not uncertain whether a policyholder will die; it is uncertain when. The random variable is age at death, and premiums are set from the probability that death occurs within the policy term.

More precisely: over a large pool of policyholders, the proportion dying in any given year is highly predictable even though no individual's date is. That is the whole basis of insurance — individual uncertainty plus aggregate regularity, which is the Central Limit Theorem doing commercial work.

4-2TheoryLevin Ex. 4-2 · p. 155

Q. "This product contains saccharin, which has been determined to cause cancer in laboratory animals." How might probability theory have played a part in this statement?

AnswerThe word "determined" hides a probability judgement at every step:
1. A sample of animals was tested, not every animal — so the finding is an inference carrying sampling error.
2. The tumour rate in the treated group was compared with a control group, and the difference was judged too large to be chance — that is a hypothesis test, and it carries a probability of being wrong.
3. Extrapolating from animals to humans, and from high doses to normal consumption, are further probabilistic judgements.
So the statement is not a certainty but a conclusion that the observed effect would be very unlikely if saccharin were harmless.

4-3TheoryLevin Ex. 4-3 · p. 155

Q. Is there really any such thing as an "uncalculated risk"? Explain.

AnswerIn one sense no, in another yes.
Whenever someone takes a risk they are making an implicit probability judgement — that the chance of loss is small enough relative to the gain. In that sense the risk is always "calculated", just subjectively and often badly.
What does not exist is a risk with no probability attached; what does exist, and commonly, is a risk whose probability has never been made explicit or checked against data. The value of probability theory is precisely in replacing an unexamined subjective estimate with a stated, defensible number.

4-4TheoryLevin Ex. 4-4 · p. 155

Q. A soft drink company alters the formula of its most popular product. How might probability theory be involved?

Answer Taste tests on a sample of consumers give \(\hat{p}\), the proportion preferring the new formula — an estimate of the population proportion with a measurable standard error.
Sample-size determination fixes how many testers are needed for the required precision (Unit 6).
Hypothesis testing decides whether the observed preference is large enough to be more than chance.
Expected value weighs the projected profit of each outcome — the new formula succeeding, failing, or alienating existing customers — against its probability (Unit 4).
This is the New Coke case: the taste tests were sound, but they measured preference in a blind sip test rather than loyalty to the brand — a reminder that a correct probability applied to the wrong question is still the wrong answer.

4-5Mutually exclusiveLevin Ex. 4-5 · p. 156

Q. Which are pairs of mutually exclusive events when drawing one card? (a) a heart and a queen; (b) a club and a red card; (c) an even number and a spade; (d) an ace and an even number. And in rolling two dice: (a) a total of 5 and a 5 on one die; (b) a total of 7 and an even number on both dice; (c) a total of 8 and an odd number on both dice; (d) a total of 9 and a 2 on one die; (e) a total of 10 and a 4 on one die.

Why: two events are mutually exclusive if no single outcome satisfies both. So for each pair, try to find one outcome that does — if you can, they are not mutually exclusive.

Why we do itWhat to write

Cards — look for a single card in both categories.

(a) Not — the queen of hearts is both.
(b) Mutually exclusive — clubs are black, so no card is both.
(c) Not — the 2 of spades is both.
(d) Mutually exclusive — an ace is rank 1, which is odd, so no card is both.

Dice — look for a single roll in both categories.

(a) Not — but note: a total of 5 cannot include a 5 (the other die would be 0), so in fact mutually exclusive.
(b) Mutually exclusive — two even numbers always sum to an even total, never 7.
(c) Not — 3+5 = 8 with both dice odd.

The last two.

(d) Mutually exclusive — a total of 9 needs (3,6),(4,5),(5,4),(6,3); none contains a 2.
(e) Not — (4,6) and (6,4) both give 10 and contain a 4.

The technique: do not reason abstractly — try to construct an outcome belonging to both. If you succeed, they are not mutually exclusive; if you can show none exists, they are.

4-6Sample spaceLevin Ex. 4-6 · p. 157

Q. A stack of nine spades numbered 2 to 10, and a die. Give a collectively exhaustive list of the possible outcomes of rolling the die and picking one card. How many elements are in the sample space?

Each outcome is a pair (die, card). Count by multiplying the choices.

Die: 6 outcomes. Card: 9 outcomes. \[ 6\times9=\mathbf{54 \text{ elements}} \]

Describe the list rather than writing all 54.

\(\{(1,2),(1,3),\dots,(1,10),(2,2),\dots,(6,10)\}\) — every pair of a die face 1–6 with a card value 2–10. The list is collectively exhaustive (it covers every possibility) and mutually exclusive (exactly one occurs).

4-7CountingLevin Ex. 4-7 · p. 157

Q. For the same die and nine-card stack, give the probability of each of these totals (die + card): 2, 3, 8, 9, 12, 14, 16.

Why: classical probability — count the (die, card) pairs giving each total and divide by 54. The die runs 1–6, the card 2–10.

Total	2	3	8	9	12	14	16
Pairs	0	1	6	6	5	3	1
P	0	1/54	6/54	6/54	5/54	3/54	1/54

Total 2 is impossible — the smallest card is 2 and the smallest die is 1, so the minimum total is 3.

\(P(2)=\mathbf{0}\); \(P(3)=1/54\) (only die 1 + card 2).

The maximum total is \(6+10=16\), reached only one way.

\(P(8)=P(9)=6/54=\mathbf{0.111}\) — these are the peak, since all six die faces can pair with a valid card.
\(P(12)=5/54\), \(P(14)=3/54\), \(P(16)=1/54=\mathbf{0.0185}\).

4-8Events & assignmentLevin Ex. 4-8 · p. 157

Q. A supporter says "chances are good" that Dheer will beat his single opponent. (a) What events could occur? (b) Is your list collectively exhaustive and mutually exclusive? (c) Knowing nothing more, what probabilities would you assign?

(a) List every possible result of a two-candidate election.

\(E_1\): Dheer wins · \(E_2\): the opponent wins · \(E_3\): a tie

(b) Test both properties.

Collectively exhaustive: yes — no other result is possible. Mutually exclusive: yes — exactly one occurs.
A two-outcome list (win / lose) would not be exhaustive unless ties are ruled out — say which you assume.

Ignoring ties: \(P(\text{Dheer})=P(\text{opponent})=\mathbf{0.5}\) each.
This is the classical approach — equal likelihood in the absence of any distinguishing information. Note the supporter's "chances are good" is a subjective probability from an interested party, so it carries no evidential weight (test 1 of the five tests for data).

4-9Card probabilitiesLevin Ex. 4-9 · p. 162

Q. Drawing one card from 52, find: (a) a seven; (b) a black card; (c) an ace or a king; (d) a black two or a black three; (e) a red face card (K, Q or J). What type of probability estimates are these?

Why we do itWhat to write

Count favourable cards over 52 each time.

(a) four 7s: \(4/52=\mathbf{0.0769}\)
(b) 26 black: \(26/52=\mathbf{0.5000}\)
(c) 4 aces + 4 kings, mutually exclusive: \(8/52=\mathbf{0.1538}\)

(d) Black twos: 2 (clubs, spades). Black threes: 2. No overlap.

\[ 4/52=\mathbf{0.0769} \]

(e) Face cards are J, Q, K — three per suit, two red suits.

\[ 6/52=\mathbf{0.1154} \] All five are classical probabilities — derived from the known structure of the deck, with no observation required.

4-11From a frequency tableLevin Ex. 4-11 · p. 163

Q. Annual commissions of 300 media salespeople: ₹0–4,999 → 15; 5,000–9,999 → 25; 10,000–14,999 → 35; 15,000–19,999 → 125; 20,000–24,999 → 70; 25,000+ → 30. Find the probability of a commission (a) between ₹5,000 and ₹10,000, (b) less than ₹15,000, (c) more than ₹20,000, (d) between ₹15,000 and ₹20,000.

Why: a relative-frequency question dressed as a frequency distribution — every answer is a class total (or sum of class totals) over 300.

(a) One class.

\[ \frac{25}{300}=\mathbf{0.0833} \]

(b) The first three classes.

\[ \frac{15+25+35}{300}=\frac{75}{300}=\mathbf{0.2500} \]

\[ \frac{70+30}{300}=\frac{100}{300}=\mathbf{0.3333} \]

(d) One class — the modal one.

\[ \frac{125}{300}=\mathbf{0.4167} \] These are relative frequency probabilities: computed from observed data, not from structure.

4-12Ratios to probabilitiesLevin Ex. 4-12 · p. 163

Q. Mr Singh's chances of getting 50–74% of his budget approved are twice those of 75–99%, and two and a half times those of 25–49%. There is no chance of less than 25% being approved, and the entire budget has been approved only once in ten years. Assign probabilities.

Why: the same "twice as likely" technique as Unit 4's Southport Autos — set the smallest to \(k\), express the others as multiples, use the one hard datum, and force the total to 1.

Why we do itWhat to write

Use the one piece of hard data first: 100% approved once in ten years.

\[ P(100\%) = \frac{1}{10}=0.10 \] and \(P(<25\%)=0\) as stated.

Express the three remaining bands in terms of one unknown. Let \(P(25\text{–}49)=k\).

\(P(50\text{–}74)=2.5k\) (2½ times the 25–49 band)
\(P(50\text{–}74)=2\times P(75\text{–}99)\) so \(P(75\text{–}99)=1.25k\)

Force the total to 1.

\[ 0 + k + 2.5k + 1.25k + 0.10 = 1 \] \[ 4.75k = 0.90 \;\Rightarrow\; k = 0.1895 \]

Back-substitute and check.

\(P(<25\%)=\mathbf{0}\) · \(P(25\text{–}49)=\mathbf{0.189}\) · \(P(50\text{–}74)=\mathbf{0.474}\) · \(P(75\text{–}99)=\mathbf{0.237}\) · \(P(100\%)=\mathbf{0.100}\)
Check: \(0+0.189+0.474+0.237+0.100=1.000\) ✓

These are subjective probabilities — based on one person's twenty years of experience, not on a repeatable experiment or a counted structure.

4-13Relative frequencyLevin Ex. 4-13 · p. 163

Q. Copier days functioning / out of service: 209/51, 217/43, 258/2, 229/31, 247/13. What is the probability of a copier being out of service?

Pool all copiers. Each has 260 days on record, so the total is \(5\times260\).

Out of service: \(51+43+2+31+13=140\)
Total copier-days: \(1300\)

Divide.

\[ P(\text{out of service})=\frac{140}{1300}=\mathbf{0.1077} \] A relative frequency probability.

Worth adding: the pooled figure hides enormous variation — copier 3 was out just 2 days while copier 1 was out 51. Reporting one average probability for "a copier" would be misleading; copier 1 should be investigated or replaced.

4-14Classify the three typesLevin Ex. 4-14 · p. 163

Q. Classify as classical, relative frequency or subjective: (a) P(Cubs win the World Series) = 0.175; (b) P(tuition increases) = 0.95; (c) P(win the lottery) = 0.00062; (d) P(a flight arrives on time) = 0.875; (e) P(two heads in two tosses) = 0.25; (f) P(car starts on a cold day) = 0.97.

Why: ask where the number came from — counted structure, observed records, or judgement.

Why we do itWhat to write

Derived from the known structure of the experiment, countable in advance.

(c) Classical — the number of winning combinations over all combinations.
(e) Classical — \(0.5\times0.5=0.25\) from the structure of a fair coin.

Derived from records of repeated past occurrences.

(d) Relative frequency — airline on-time records.
(f) Relative frequency — the owner's own experience over many cold mornings.

A one-off event that cannot be repeated, so only judgement is available.

(a) Subjective — this season happens once.
(b) Subjective — next year's tuition decision happens once.

Matches Levin's key (SC 4-4 uses the same three-way split). The discriminator for (a) and (f): both feel like judgement, but a car starting has been observed many times, while a specific season has not.

SC 4-1 / 4-2Two diceLevin SC 4-1, 4-2 · p. 157

Q. Write the sample space for rolling two dice, then give P(1), P(2), P(5), P(6), P(7), P(10), P(11) for the total.

The sample space is all ordered pairs — 36 of them, each equally likely.

\(\{(1,1),(1,2),\dots,(6,6)\}\), \(6\times6=36\) outcomes.

Count the pairs giving each total.

\(P(1)=\mathbf{0/36}\) — impossible, the minimum total is 2
\(P(2)=1/36\) · \(P(5)=4/36\) · \(P(6)=5/36\) · \(P(7)=\mathbf{6/36}\) (the peak)
\(P(10)=3/36\) · \(P(11)=2/36\)

Matches Levin's printed answers exactly. Note the ordered pairs: (2,3) and (3,2) are different outcomes. Treating them as one is the classic error that makes the 36 become 21 and every probability wrong.

Topic 2 of 4 · Levin §4.4 · pp. 168–170 · 6 textbook + 6 exam-shaped

The Addition Rule — "A or B"

One formula, and one question to ask before using it: can both happen at once? If they can, you have double-counted the overlap and must subtract it. That subtraction is the entire topic.

How to know a question belongs to this topic

The single tell

The word OR — or "either", or "at least one of two things".

Give-away words

A or B either … or … damaged or overripe disk-drive and/or keyboard failure P(A or B) = Venn diagram

The one decision

Can A and B both happen?
No → mutually exclusive → \(P(A\text{ or }B)=P(A)+P(B)\).
Yes → \(P(A\text{ or }B)=P(A)+P(B)-P(AB)\) — subtract the overlap, or you count it twice.

Reading a Venn diagram

The numbers inside the overlap belong to both events. So \(P(A)\) is the whole of circle A — its own region plus the overlap. Forgetting to include the overlap in \(P(A)\) is the standard Venn error.

Don't confuse it with

If the question…	then…
says "both" or "and"	Topic 3 — multiplication
says "at least one" of many events	Topic 3 — use \(1-P(\text{none})\)
says "given that"	Topic 3 or 4 — conditional

Master Method — Topic 2

Formula bank

Addition rule
(general)

\[ P(A \text{ or } B) = P(A) + P(B) - P(AB) \]

Mutually exclusive

\[ P(A \text{ or } B) = P(A) + P(B) \qquad\text{since } P(AB)=0 \]

Three events

\[ P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC) \]

Complement

\[ P(\text{neither } A \text{ nor } B) = 1 - P(A \text{ or } B) \]

Traps

Forgetting to subtract \(P(AB)\). Gives a probability that can exceed 1 — an instant sign you have erred.
Assuming mutually exclusive without checking. Say why: "a rod can have both flaws, so they are not mutually exclusive."
On a Venn diagram, reading only the exclusive region as \(P(A)\). Circle A includes the overlap.
"And/or" means OR. "Resistant to disk-drive and/or keyboard failure" is the union.

Textbook bank §4.4, pp. 168–170 · solved

6 exercises Levin §4.4 Probability rules pp. 164–170

4-15Venn diagramLevin Ex. 4-15 · p. 169

Q. From a Venn diagram with 50 total outcomes, region A only = 11, the overlap = 7, region B only = some value: give P(A), P(B), P(A or B).

Why: the method matters more than the specific numbers. Circle A = its own region plus the overlap; the same for B. Then apply the addition rule — and note the overlap is exactly what you subtract.

Why we do itWhat to write

Read each circle as its exclusive region + the overlap.

\[ P(A)=\frac{(A\text{ only})+(\text{overlap})}{50} \qquad P(B)=\frac{(B\text{ only})+(\text{overlap})}{50} \]

With A only = 11 and overlap = 7:

\[ P(A)=\frac{11+7}{50}=\frac{18}{50}=\mathbf{0.36} \qquad P(AB)=\frac{7}{50}=\mathbf{0.14} \]

Union — and the cross-check that makes Venn questions safe.

\[ P(A\text{ or }B)=P(A)+P(B)-P(AB) \] Equivalently, just count every outcome inside either circle and divide by 50 — the two routes must agree. If they do not, you have mis-assigned the overlap.

Levin's SC 4-5 is the same diagram fully specified: 50 total, A only = 8, overlap = 6, B only = 13, outside = 23. Then \(P(A)=14/50=0.28\), \(P(B)=19/50=0.38\), \(P(A\text{ or }B)=(8+6+13)/50=27/50=0.54\) — and the formula gives \(0.28+0.38-0.12=0.54\) ✓.

4-16Three-circle VennLevin Ex. 4-16 · p. 169

Q. From a three-circle Venn diagram with 100 total outcomes and regions 10, 2, 20, 3, 6, 4, 25, give P(A), P(B), P(C), P(A or B), P(A or C), and P(B but not (A or C)).

Why: with three circles there are seven regions. Add every region inside a circle to get its probability; for a union, add every region inside either; for "B but not A or C", take only B's exclusive region.

Why we do itWhat to write

A circle's probability is the sum of every region lying inside it — exclusive part plus all its overlaps.

\[ P(A)=\frac{\text{all regions inside }A}{100} \] and likewise for B and C.

A union is every region inside either circle, counted once.

\[ P(A\text{ or }B)=\frac{\text{regions inside }A\text{ or }B}{100} \] — counting regions directly avoids the three-event inclusion–exclusion formula entirely.

"B but not (A or C)" is B's exclusive region alone.

\[ P(B\text{ but not }(A\text{ or }C))=\frac{\text{B's exclusive region}}{100} \] With B's exclusive region = 20: \(\mathbf{0.20}\).

Technique worth keeping: on any Venn question, count regions rather than applying inclusion–exclusion. It is faster and cannot go wrong on the signs — the formula is only needed when you are given probabilities instead of a diagram.

4-17Derive the ruleLevin Ex. 4-17 · p. 169

Q. (a) What can you say about \(P(AB)\) when A and B are mutually exclusive? (b) Develop an expression for \(P(A\text{ or }B\text{ or }C)\) without assuming mutual exclusivity. (c) Rewrite it if A and B are mutually exclusive but A–C and B–C are not. (d) Rewrite it if A–B and A–C are mutually exclusive but B–C is not.

Why we do itWhat to write

(a) Mutually exclusive means they cannot occur together.

\[ P(AB)=\mathbf{0} \] which is why the general rule collapses to simple addition.

(b) Inclusion–exclusion: add the singles, subtract every pair, add back the triple (which was subtracted three times having been added three times).

\[ P(A\cup B\cup C)=P(A)+P(B)+P(C) \] \[ -P(AB)-P(AC)-P(BC)+P(ABC) \]

\[ P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(AC)-P(BC) \] \(P(ABC)=0\) too, since it requires \(AB\).

(d) Now \(P(AB)=P(AC)=0\), so \(P(ABC)=0\) as well; only \(P(BC)\) survives.

\[ P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(BC) \]

The principle to state: every mutually exclusive pair kills its own intersection term and any higher term containing it. That is why these expressions shorten so quickly.

4-19Solve backwardsLevin Ex. 4-19 · p. 170

Q. Disk-drive failures occur one-third as often as keyboard failures. \(P(\text{both}) = 0.05\). If the computer is 80% resistant to disk-drive and/or keyboard failure, how low must the disk-drive failure probability be?

Why: "80% resistant" means the union has probability 0.20. Put the addition rule in terms of one unknown and solve — and note "and/or" is the union.

Why we do itWhat to write

Translate "80% resistant" into the union.

\[ P(D \text{ or } K)=1-0.80=0.20 \]

Express \(K\) in terms of \(D\): disk-drive failures are one-third as often, so \(P(D)=\tfrac13 P(K)\), i.e. \(P(K)=3P(D)\).

Let \(P(D)=d\), then \(P(K)=3d\).

Substitute into the addition rule.

\[ d + 3d - 0.05 = 0.20 \] \[ 4d = 0.25 \]

Solve and check.

\[ P(D)=\mathbf{0.0625} \qquad P(K)=0.1875 \] Check: \(0.0625+0.1875-0.05=0.20\) ✓

4-20OverlapLevin Ex. 4-20 · p. 170

Q. Of every 1,000 fuel rods inspected, 10 have interior flaws, 8 have casing flaws, and 5 have both. What is the probability of flaws in a fuel rod?

Why: "flaws" means either kind — a union. The 5 with both are already counted inside the 10 and inside the 8, so they must be subtracted once.

Write the three given probabilities.

\[ P(I)=\tfrac{10}{1000}=0.010,\quad P(C)=\tfrac{8}{1000}=0.008,\quad P(IC)=\tfrac{5}{1000}=0.005 \]

Addition rule — they are clearly not mutually exclusive, since 5 rods have both.

\[ P(I\text{ or }C)=0.010+0.008-0.005=\mathbf{0.013} \] i.e. 13 flawed rods per 1,000 — not 18, which is what adding without subtracting would give.

The sanity check: count directly — 5 have interior only, 3 have casing only, 5 have both = 13 flawed rods. Matches. Any union answer can be verified this way when raw counts are given.

SC 4-6Compare two unionsLevin SC 4-6 · p. 170

Q. Station 1: P(pump failure) = 0.07, P(leakage) = 0.10, P(both) = 0. Station 2: P(pump failure) = 0.09, P(leakage) = 0.12, P(both) = 0.06. Which station has the higher probability of being shut down, and which should Samreen take over?

Why: a shutdown happens on either failure, so compute each union. The twist: Station 2 has larger individual probabilities but a large overlap, which makes its union smaller.

Why we do itWhat to write

Station 1 — the two failures are mutually exclusive, so nothing to subtract.

\[ P_1 = 0.07+0.10-0 = \mathbf{0.17} \]

Station 2 — they overlap, so subtract.

\[ P_2 = 0.09+0.12-0.06 = \mathbf{0.15} \]

Compare and answer the business question.

Station 1 has the higher shutdown probability (0.17 vs 0.15), so Samreen should take over Station 2.
Counter-intuitive but correct: Station 2's failures are worse individually, but they tend to happen together, so they cause fewer separate shutdowns.

Matches Levin's printed answer exactly. This is the best question in the topic — it shows why the overlap term is not bookkeeping but a real effect: correlated failures shut a plant down less often than independent ones.

Six more the addition rule in the shapes an exam uses

Ask "can both happen?" before writing anything.

X1Basic union

Q. \(P(A) = 0.45\), \(P(B) = 0.35\), \(P(AB) = 0.15\). Find \(P(A\text{ or }B)\) and \(P(\text{neither})\).

Solution

Addition rule, then complement.

\[ P(A\text{ or }B)=0.45+0.35-0.15=\mathbf{0.65} \] \[ P(\text{neither})=1-0.65=\mathbf{0.35} \]

X2Mutually exclusive?

Q. A card is drawn. Find P(a king or a heart), and explain why the answer is not 4/52 + 13/52.

Solution

The king of hearts is both, so they are not mutually exclusive.

\[ P(K)=\tfrac{4}{52},\quad P(H)=\tfrac{13}{52},\quad P(KH)=\tfrac{1}{52} \]

Subtract the one card counted twice.

\[ P=\frac{4+13-1}{52}=\frac{16}{52}=\mathbf{0.3077} \] Adding blindly gives 17/52 — it counts the king of hearts twice.

X3From counts

Q. Of 200 customers, 120 bought product A, 90 bought product B, and 40 bought both. Find P(bought A or B) and P(bought neither).

Solution

Addition rule on counts.

\[ P(A\text{ or }B)=\frac{120+90-40}{200}=\frac{170}{200}=\mathbf{0.85} \]

Complement, and verify by direct counting.

\[ P(\text{neither})=1-0.85=\mathbf{0.15} \;(30\text{ customers}) \] Check: A only 80, B only 50, both 40, neither 30 → 200 ✓

X4Solve backwards

Q. \(P(A) = 0.5\), \(P(A\text{ or }B) = 0.7\), and A and B are mutually exclusive. Find \(P(B)\). What if instead \(P(AB) = 0.1\)?

Solution

Mutually exclusive ⇒ nothing to subtract.

\[ 0.7=0.5+P(B) \;\Rightarrow\; P(B)=\mathbf{0.2} \]

With an overlap, the same union needs a larger \(P(B)\).

\[ 0.7=0.5+P(B)-0.1 \;\Rightarrow\; P(B)=\mathbf{0.3} \] Overlapping events must individually be larger to produce the same union.

X5Three events

Q. A machine can fail from wear (0.10), overheating (0.08) or power surge (0.05). No two causes ever occur together. Find P(the machine fails) and P(it does not).

Solution

All pairwise intersections are zero, so simply add.

\[ P(\text{fail})=0.10+0.08+0.05=\mathbf{0.23} \] \[ P(\text{no failure})=1-0.23=\mathbf{0.77} \]

X6Impossible probabilities

Q. Someone reports \(P(A) = 0.6\), \(P(B) = 0.7\) and \(P(A\text{ or }B) = 0.5\). Show that this is impossible.

Solution

Why: a union can never be smaller than either of its parts — \(A\) alone already guarantees \(A\) or \(B\).

The union must be at least as large as the larger event.

\(P(A\text{ or }B) \ge \max(0.6,\,0.7)=0.7\), but 0.5 was reported. Impossible.

Confirm via the formula.

\[ P(AB)=P(A)+P(B)-P(A\text{ or }B)=0.6+0.7-0.5=0.8 \] But \(P(AB)\) cannot exceed \(P(A)=0.6\). Two independent contradictions — the figures are inconsistent.

Topic 3 of 4 · Levin §4.5–§4.6 · pp. 176–187 · 19 textbook questions

Multiplication, Conditional Probability and Independence

"A and B". One formula in two versions, and the whole difficulty is deciding which — does knowing that A happened change the probability of B? Get that right and the arithmetic is a product.

How to know a question belongs to this topic

The single tell

The word AND, or "given that" where the condition comes first in time. Also every "at least one of several" question.

Give-away words

both … and … given that the first was … with replacement without replacement independently in succession at least one of the four

The one decision

Independent (with replacement, separate machines, separate people): \(P(AB)=P(A)\times P(B)\) and \(P(B|A)=P(B)\).
Dependent (without replacement, or a contingency table): \(P(AB)=P(A|B)\times P(B)\) and \(P(B|A)=\dfrac{P(AB)}{P(A)}\).

The "with replacement" switch

This single phrase decides everything. With replacement → independent, probabilities never change. Without → dependent, and the denominator shrinks on each draw.

"At least one" — always the complement

\(P(\text{at least one}) = 1 - P(\text{none})\), and \(P(\text{none})\) is a product of the individual failure probabilities. Never sum the cases.

Don't confuse it with

If the question…	then…
asks "A or B"	Topic 2
asks P(cause \| effect) — reversing the given order	Topic 4 — Bayes
asks P(effect \| cause) — the natural order	Here

Master Method — Topic 3

Formula bank

Joint, independent

\[ P(AB) = P(A)\times P(B) \]

Joint, dependent

\[ P(AB) = P(A|B)\times P(B) = P(B|A)\times P(A) \]

Conditional, dependent

\[ P(B|A) = \frac{P(AB)}{P(A)} \]

Conditional, independent

\[ P(B|A) = P(B) \qquad\text{knowing } A \text{ tells you nothing} \]

At least one

\[ P(\text{at least one}) = 1 - P(\text{none}) = 1 - \prod_i (1-p_i) \]

Test for independence

\[ A,B \text{ independent} \iff P(AB)=P(A)P(B) \iff P(B|A)=P(B) \]

The routine

Decide independent or dependent, and say which. "With replacement, so the draws are independent."
Write the events with symbols — \(M\) = male, \(F\) = first-time offender. It halves the confusion in table questions.
Pick the version of the formula that matches, and substitute.
For "at least one", switch to the complement before doing anything else.
Sanity-check: a conditional probability must lie between 0 and 1, and a joint probability can never exceed either of its marginals.

Traps

Multiplying dependent probabilities as if independent. Check for "without replacement".
Inverting the conditional. \(P(A|B) \ne P(B|A)\) unless \(P(A)=P(B)\). That confusion is the whole reason Bayes exists.
Summing for "at least one". Double-counts the overlaps; use the complement.
Reading a contingency table down instead of across. The denominator of \(P(B|A)\) is the total for \(A\), not the grand total.

Textbook bank §4.5 independence, §4.6 dependence · pp. 176–187 · solved

19 exercises Levin §4.5 pp. 171–179 · §4.6 pp. 179–187

Say "independent" or "dependent" out loud before you write.

4-21IndependenceLevin Ex. 4-21 · p. 177

Q. What is the probability that a couple's second child is (a) a boy, given the first was a girl? (b) a girl, given the first was a girl?

Solution

Why: the sexes of successive children are independent — the second birth has no memory of the first.

Independent ⇒ the condition is irrelevant.

(a) \(P(\text{boy}|\text{girl first})=P(\text{boy})=\mathbf{0.5}\)
(b) \(P(\text{girl}|\text{girl first})=P(\text{girl})=\mathbf{0.5}\)
This is the gambler's fallacy in miniature — a run of girls does not make a boy "due".

4-22Successive rollsLevin Ex. 4-22 · p. 177

Q. Rolling two dice, find the probability of (a) a total of 7 then a total of 11; (b) a total of 21 on the first two rolls combined; (c) a total of 6 on the first three rolls combined.

Solution

Why: rolls are independent, so joint probabilities multiply. Parts (b) and (c) need you to count the combinations of roll totals, not dice faces.

(a) \(P(7)=6/36\), \(P(11)=2/36\); multiply.

\[ \frac{6}{36}\times\frac{2}{36}=\frac{1}{6}\times\frac{1}{18}=\frac{1}{108}=\mathbf{0.00926} \]

(b) Ways to make 21 from two roll-totals: (9,12), (10,11), (11,10), (12,9). Multiply the ways for each.

\[ 4(1)+3(2)+2(3)+1(4)=20 \text{ ways out of } 36^2=1296 \] \[ =\frac{20}{1296}=\frac{5}{324}=\mathbf{0.01543} \]

\[ \left(\frac{1}{36}\right)^3=\frac{1}{46656}=\mathbf{0.0000214} \] Only one way exists — (2,2,2).

4-23With replacementLevin Ex. 4-23 · p. 177

Q. A bag has 32 marbles: 4 red, 9 black, 12 blue, 6 yellow, 1 purple. Drawn one at a time with replacement. Find P(second is yellow | first was yellow), P(second is yellow | first was black), P(third is purple | first two were purple).

Solution

Why: "with replacement" is the whole question — the bag is restored every time, so every draw is independent and the conditions are irrelevant.

Independent ⇒ ignore every condition.

(a) \(P(\text{yellow})=\dfrac{6}{32}=\mathbf{0.1875}\)
(b) \(P(\text{yellow})=\dfrac{6}{32}=\mathbf{0.1875}\) — identical to (a)
(c) \(P(\text{purple})=\dfrac{1}{32}=\mathbf{0.03125}\)

If it were WITHOUT replacement, (a) would be \(5/31 = 0.161\) and (c) would be impossible (only one purple marble exists). That contrast is exactly what the question is testing.

4-24Sequential gameLevin Ex. 4-24 · p. 177

Q. Four men draw from balls 1–4; whoever draws 4 loses. The remaining three redraw; whoever draws 3 loses. The last two redraw; whoever draws 1 wins. (a) P(John does not lose in the first two draws)? (b) P(Paul wins)?

Solution

(a) Round 1: 4 men, one loses ⇒ P(survive) = 3/4. Round 2: 3 men left, one loses ⇒ P(survive) = 2/3.

\[ P=\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}=\mathbf{0.5} \]

(b) The game treats all four men identically — no one has an advantage.

By symmetry, \(P(\text{Paul wins})=\mathbf{1/4}=0.25\).
Confirm the long way: \(\tfrac34\times\tfrac23\times\tfrac12=\tfrac14\) ✓ — survive both eliminations, then win the final draw.

4-25Independent inspectorsLevin Ex. 4-25 · p. 177

Q. Two independent inspections; a restaurant passes only if both pass it. Inspector A passes 2% of restaurants with violations, B passes 7%. Find (a) P(A passes | B found a violation), (b) P(B passes a violator | A passes it), (c) P(a restaurant with a violation is passed).

Solution

Why: "independent" is stated, so conditions do nothing to (a) and (b). Only (c) needs a multiplication.

(a) and (b): independence means each inspector's behaviour is unaffected by the other.

(a) \(P(A\text{ passes})=\mathbf{0.02}\)
(b) \(P(B\text{ passes})=\mathbf{0.07}\)

\[ P=0.02\times0.07=\mathbf{0.0014} \] Only 14 violators in 10,000 get through — the double inspection is very effective, precisely because the two inspectors are independent.

4-26Check a claimLevin Ex. 4-26 · p. 178

Q. Four floodgates each fail independently 4% of the time. (a) If gate 1 is out, what is the probability gates 2 and 3 are out? (b) You are told the chance of all four being out is less than 1 in 5,000,000. Is that true?

Solution

(a) Independent ⇒ gate 1's state is irrelevant; just multiply gates 2 and 3.

\[ P=0.04\times0.04=\mathbf{0.0016} \]

(b) All four out.

\[ P=(0.04)^4=0.00000256 \]

Convert to "1 in n" and compare.

\[ \frac{1}{0.00000256}=390{,}625 \] So the true chance is about 1 in 390,000 — the claim of "less than 1 in 5,000,000" is FALSE. The real risk is roughly 13 times higher than stated.

4-27Chain of approvalsLevin Ex. 4-27 · p. 178

Q. A report must be approved by the group leader (85%), department head (80%) and division chief (82%), acting independently and in that order. What is the probability the report reaches the Ministry?

Solution

All three must approve — independent, so multiply.

\[ P=0.85\times0.80\times0.82=\mathbf{0.5576} \] Only about 56% of reports get through, even though every individual approval rate is above 80%.

The lesson worth stating: chains of independent approvals multiply, so success decays fast. Three 80% gates give 51%; five give 33%. This is why long approval chains are a design problem, not just an inconvenience.

4-28Contingency tableLevin Ex. 4-28 · p. 178

Q. Orange juice cartons sold, 30 days each for morning / afternoon / evening:

Number sold	Morning	Afternoon	Evening
0–19	3	8	2
20–39	3	4	3
40–59	12	6	4
60–79	4	9	9
80–99	5	3	6
100 or more	3	0	6
Total	30	30	30

Find (a) P(evening and 80–99), (b) P(39 or fewer and afternoon), (c) P(0–19 or 100+).

Solution

Why: the grand total is \(3\times30=90\). Every "and" is a single cell (or a group of cells in one column); every "or" is a sum across the table.

(a) A single cell — evening column, 80–99 row.

\[ P=\frac{6}{90}=\mathbf{0.0667} \]

(b) "39 or fewer" spans two rows, restricted to the afternoon column.

\[ P=\frac{8+4}{90}=\frac{12}{90}=\mathbf{0.1333} \]

\[ P=\frac{(3+8+2)+(3+0+6)}{90}=\frac{22}{90}=\mathbf{0.2444} \]

The trap in (b): "during a randomly selected afternoon" could be read as given it is an afternoon, which would give \(12/30 = 0.40\). Levin intends the joint probability over all 90 days. State your reading — either is defensible if you say which.

4-29At least oneLevin Ex. 4-29 · p. 178

Q. Four billboards are noticed by a motorist with probabilities 0.75, 0.82, 0.87 and 0.90, independently. What is the probability that a motorist notices at least one?

Solution

Why: "at least one" of four events has 15 favourable cases; its complement has exactly one. Always take the complement.

P(misses each one) is one minus its notice probability.

\[ 0.25,\; 0.18,\; 0.13,\; 0.10 \]

Independent ⇒ multiply for "misses all four".

\[ P(\text{none})=0.25\times0.18\times0.13\times0.10=0.000585 \]

Complement.

\[ P(\text{at least one})=1-0.000585=\mathbf{0.999415} \] Over 99.9% — the campaign reaches essentially every motorist.

4-30Four-part conditionalLevin Ex. 4-30 · p. 186

Q. A and B are dependent, \(P(A)=0.39\), \(P(B)=0.21\), \(P(A\text{ or }B)=0.47\). Find (a) P(neither), (b) P(both), (c) P(B|A), (d) P(A|B).

Solution

Why: get \(P(AB)\) first by rearranging the addition rule — everything else follows from it.

(a) Complement of the union.

\[ P(\text{neither})=1-0.47=\mathbf{0.53} \]

(b) Rearrange \(P(A\text{ or }B)=P(A)+P(B)-P(AB)\).

\[ P(AB)=0.39+0.21-0.47=\mathbf{0.13} \]

\[ P(B|A)=\frac{0.13}{0.39}=\mathbf{0.3333} \]

(d) Same joint, different denominator.

\[ P(A|B)=\frac{0.13}{0.21}=\mathbf{0.6190} \] Note \(P(B|A)\ne P(A|B)\) — 0.33 vs 0.62 from the same joint probability. That asymmetry is what Topic 4 is about.

4-32ConsistencyLevin Ex. 4-32 · p. 186

Q. \(P(A)=0.65\), \(P(B)=0.80\), \(P(A|B)=P(A)\), and \(P(B|A)=0.85\). Is this a consistent assignment?

Solution

Why: \(P(A|B)=P(A)\) declares independence. Independence is symmetric — so it forces \(P(B|A)=P(B)\) too. Check whether it does.

\(P(A|B)=P(A)\) means A and B are independent.

Independence ⇒ \(P(B|A)\) must equal \(P(B)=0.80\).

But the question states 0.85.

Not consistent. \(P(B|A)=0.85 \ne 0.80=P(B)\).
Confirm via the joint: independence gives \(P(AB)=0.65(0.80)=0.52\), but \(P(B|A)=0.85\) gives \(P(AB)=0.85(0.65)=0.5525\). Two different values for the same quantity — a contradiction.

4-33Conditional from percentagesLevin Ex. 4-33 · p. 186

Q. Of soup-kitchen visitors, 59% are men, 32% are alcoholics, and 21% are male alcoholics. What is the probability that a random male visitor is an alcoholic?

Solution

Identify the joint and the condition. "Male alcoholics" is the joint; "a male visitor" is the condition.

\(P(M)=0.59\), \(P(A)=0.32\), \(P(AM)=0.21\)

Divide by the probability of the given event.

\[ P(A|M)=\frac{P(AM)}{P(M)}=\frac{0.21}{0.59}=\mathbf{0.3559} \] Higher than the overall 32%, so being male is associated with alcoholism here — the events are not independent.

4-34Both directionsLevin Ex. 4-34 · p. 186

Q. 60% of accidents occur at night, 52% are alcohol-related, 37% are both. Find (a) P(alcohol | night), (b) P(night | alcohol).

Solution

Why: the same joint probability, two different denominators — the clearest demonstration that conditionals are not symmetric.

(a) Condition on night ⇒ divide by \(P(N)\).

\[ P(A|N)=\frac{0.37}{0.60}=\mathbf{0.6167} \]

(b) Condition on alcohol ⇒ divide by \(P(A)\).

\[ P(N|A)=\frac{0.37}{0.52}=\mathbf{0.7115} \] Same numerator, different answers. 62% of night accidents involve alcohol; 71% of alcohol accidents happen at night.

4-35Joint, dependentLevin Ex. 4-35 · p. 186

Q. If a hurricane forms in the eastern Bay of Bengal there is a 76% chance it strikes Maharashtra. The probability of a hurricane forming there in a year is 0.85. What is the probability that a hurricane forms and strikes Maharashtra this year?

Solution

Joint = conditional × marginal of the condition.

\[ P(FS)=P(S|F)\times P(F)=0.76\times0.85=\mathbf{0.646} \]

4-36Joint, dependentLevin Ex. 4-36 · p. 187

Q. P(favourable evaluation) = 0.65. P(upgrade | favourable) = 0.85. What is P(upgrade and favourable)?

Solution

Same structure as 4-35.

\[ P(UF)=0.85\times0.65=\mathbf{0.5525} \]

4-38Both directionsLevin Ex. 4-38 · p. 187

Q. P(pilots strike) = 0.75, P(drivers strike) = 0.65, and if the drivers strike there is a 90% chance the pilots strike in sympathy. Find (a) P(both strike), (b) P(drivers strike | pilots strike).

Solution

(a) Joint from the conditional given.

\[ P(PD)=P(P|D)\times P(D)=0.90\times0.65=\mathbf{0.585} \]

(b) Reverse the conditional using the same joint.

\[ P(D|P)=\frac{P(PD)}{P(P)}=\frac{0.585}{0.75}=\mathbf{0.78} \]

Note what part (b) is: reversing \(P(P|D)=0.90\) into \(P(D|P)=0.78\) is Bayes' theorem in its simplest two-event form. Topic 4 just does the same thing when there are three or more possible causes.

SC 4-7With replacementLevin SC 4-7 · p. 176

Q. Two cards drawn one at a time with replacement. Find P(second is a face card | first was red), P(second is an ace | first was a face card), P(second is a black jack | first was a red ace).

Solution

With replacement ⇒ independent ⇒ conditions irrelevant.

(a) \(12/52=\mathbf{0.2308}\) — 12 face cards
(b) \(4/52=\mathbf{0.0769}\) — 4 aces
(c) \(2/52=\mathbf{0.0385}\) — 2 black jacks

SC 4-10Contingency tableLevin SC 4-10 · p. 188

Q. 250 shoplifters classified male/female (M/W) and first-time/repeat (F/R): male first-time 60, male repeat 70, female first-time 44, female repeat 76. Find P(M), P(F|M), P(W|R), P(W|F), P(M and R).

Solution

Why: the model contingency-table question. Marginal = row or column total ÷ 250. Conditional = cell ÷ the total of the given category.

Marginal — the whole male row.

\[ P(M)=\frac{60+70}{250}=\frac{130}{250}=\mathbf{0.520} \]

Conditional on male — denominator is the male total, not 250.

\[ P(F|M)=\frac{60/250}{130/250}=\frac{60}{130}=\mathbf{0.462} \]

Conditional on repeat — denominator is the repeat total \(70+76=146\).

\[ P(W|R)=\frac{76}{146}=\mathbf{0.521} \]

Conditional on first-time — denominator \(60+44=104\).

\[ P(W|F)=\frac{44}{104}=\mathbf{0.423} \]

Joint — a single cell over the grand total.

\[ P(M\text{ and }R)=\frac{70}{250}=\mathbf{0.280} \]

All five match Levin's printed answers exactly. The rule to carry away: a joint probability divides by the grand total; a conditional divides by the total of the given category.

SC 4-9Joint from conditionalLevin SC 4-9 · p. 185

Q. P(two cars | income > ₹6,00,000) = 0.75. 60% of households had incomes over ₹6,00,000 and 52% had two cars. What is P(two cars and income over ₹6,00,000)?

Solution

Joint = conditional × marginal of the condition. The 52% is not needed here.

\[ P(C \text{ and } I)=P(C|I)\times P(I)=0.75\times0.60=\mathbf{0.45} \]

The 52% is a distractor — but a useful one: with it you can also get \(P(I|C)=0.45/0.52=0.865\), the reverse conditional. Questions often supply the extra marginal precisely so a later part can ask for the reversal.

Topic 4 of 4 · Levin §4.7 · pp. 194–196 · Q5 (a) — 5 marks in both papers

Bayes' Theorem — Reversing a Conditional

The highest-value single question in the unit, and the most mechanical once you see it as a table rather than a formula. Four columns, one total row, and the answer is a ratio. Both papers ask it on a classifier — a test, a screen, a detector — and both want a managerial sentence at the end.

How to know a question belongs to this topic

The single tell

You are given \(P(\text{evidence} \mid \text{cause})\) and asked for \(P(\text{cause} \mid \text{evidence})\) — the conditional runs backwards. The effect has already been observed and you must work out which cause produced it.

Give-away words

given that it was collected, what is the probability it was called on which city did it most likely choose if the picnic was cancelled, what is the probability… a randomly selected player WITH an injury revise the prior posterior probability

The structural signature

The question lists two or more mutually exclusive causes with probabilities that sum to 1 (the priors), then a conditional probability of the same piece of evidence under each (the likelihoods). If you can see those two lists, it is Bayes.

What a full answer contains

The four-column table · the total \(P(\text{evidence})\) written out as the law of total probability · the posterior as a ratio · and a sentence in the story's language. The Final's key awards a mark for that last sentence.

Don't confuse it with

If the question…	then…
asks P(effect \| cause) — the natural, forward direction	Topic 3 — plain conditional
asks P(cause \| effect) — backwards	Here
gives a contingency table of raw counts	Topic 3 — just divide cell by row total
gives priors and likelihoods separately	Here

Master Method — Topic 4

Formula bank

Law of total
probability

\[ P(E) = \sum_i P(A_i)\,P(E \mid A_i) \]

Bayes' theorem

\[ P(A_i \mid E) = \frac{P(A_i)\,P(E \mid A_i)}{P(E)} = \frac{P(A_i)\,P(E\mid A_i)}{\sum_j P(A_j)P(E\mid A_j)} \]

\(A_i\) the possible causes — mutually exclusive, exhaustive

\(P(A_i)\) the prior — before the evidence

\(P(E|A_i)\) the likelihood — chance of the evidence if that cause holds

\(P(A_i|E)\) the posterior — revised, after the evidence

You never need to remember the formula if you remember the table: prior × likelihood = joint, then each joint ÷ the total of the joints. The total column is the law of total probability, and the division is Bayes.

The five-line method — this is the whole question

STEP 1 List the CAUSES. Check their priors sum to 1. STEP 2 Write the LIKELIHOOD of the observed evidence under each cause. STEP 3 Build the table — one row per cause: Cause P(cause) P(E | cause) JOINT = product POSTERIOR ───────────────────────────────────────────────────────────────────────── A₁ 0.40 0.45 0.1800 0.1800/T A₂ 0.35 0.60 0.2100 0.2100/T A₃ 0.25 0.35 0.0875 0.0875/T ───────────────────────────────────────────────────────────────────────── TOTAL T = 0.4775 ← this is P(E) STEP 4 Divide each joint by T. The posteriors must sum to 1 — CHECK IT. STEP 5 Say what it means in the story's own words.

How the exam asks it

Paper	The question
Final	Q5 (a), 5 marks — Bayes on a classifier: total probability, then the posterior, then a managerial insight QP p. 3 · worked here
Re-Exam	Q5 (a), 5 marks — structurally identical, different story QP p. 3 · worked here
Class notes	The heart-attack and rare-disease examples, worked in the same five lines notes pp. 1–2

Traps

Inverting the conditional by mistake. \(P(\text{disease}|\text{positive})\) is not the test's accuracy. This is the base-rate fallacy, and it is what the question is testing.
Forgetting a cause. The priors must sum to 1. If two are given and they total 0.85, there is a third at 0.15.
Using the wrong denominator. Divide by the total of the joints, not by the prior or the likelihood.
Stopping at the number. Both papers award a mark for the managerial reading.
Not checking the posteriors sum to 1. One line, catches nearly every arithmetic slip.

Textbook bank — every Levin Bayes exercise §4.7, pp. 194–196 · all 10, solved as tables

10 exercises Levin §4.7 Revising prior estimates: Bayes' theorem pp. 188–196

4-41The bare structureLevin Ex. 4-41 · p. 194

Q. Outcomes A, B, C are mutually exclusive with \(P(A)=0.2\), \(P(B)=0.65\). A second experiment gives X or Y, with \(P(X|A)=0.75\), \(P(X|B)=0.60\), \(P(X|C)=0.40\). Find \(P(A|X)\), \(P(B|X)\), \(P(C|X)\), and \(P(Y)\).

Why: the method with the story stripped away. Note the first move: \(P(C)\) is not given and must be recovered from the fact that the priors sum to 1.

Why we do itWhat to write

Recover the missing prior — the causes are exhaustive.

\[ P(C)=1-0.20-0.65=\mathbf{0.15} \]

Build the table: prior × likelihood = joint.

\(A:\ 0.20(0.75)=0.1500\)
\(B:\ 0.65(0.60)=0.3900\)
\(C:\ 0.15(0.40)=0.0600\)

Total the joints — this is \(P(X)\), the law of total probability.

\[ P(X)=0.1500+0.3900+0.0600=\mathbf{0.60} \]

Divide each joint by the total.

\[ P(A|X)=\tfrac{0.15}{0.60}=\mathbf{0.25} \quad P(B|X)=\tfrac{0.39}{0.60}=\mathbf{0.65} \quad P(C|X)=\tfrac{0.06}{0.60}=\mathbf{0.10} \] Check: \(0.25+0.65+0.10=1.00\) ✓

\(Y\) is the complement of \(X\).

\[ P(Y)=1-P(X)=\mathbf{0.40} \]

Worth noticing: \(P(B)=0.65\) and \(P(B|X)=0.65\) — the evidence \(X\) left B's probability unchanged, because B's likelihood (0.60) happens to equal \(P(X)\). Evidence only shifts a belief when it is more or less likely under that cause than overall.

4-42Three causesLevin Ex. 4-42 · p. 194

Q. 70% of delinquent accounts are called on personally, 20% phoned, 10% sent a letter. The chances of collecting are 0.75, 0.60 and 0.65 respectively. A payment has just been received. What is the probability the account (a) was called on personally, (b) received a phone call, (c) received a letter?

Why: the evidence is "payment received"; the causes are the three methods. Classic backwards conditional.

Method	Prior P(M)	P(collect \| M)	Joint	Posterior
Personal call	0.70	0.75	0.5250	0.7394
Phone	0.20	0.60	0.1200	0.1690
Letter	0.10	0.65	0.0650	0.0915
Total	1.00	—	0.7100	0.9999 ✓

The total of the joints is the overall collection rate.

\[ P(\text{collect})=0.525+0.120+0.065=\mathbf{0.71} \]

Each posterior is its joint over that total.

\[ \text{(a) } \frac{0.525}{0.71}=\mathbf{0.7394} \quad \text{(b) } \frac{0.120}{0.71}=\mathbf{0.1690} \quad \text{(c) } \frac{0.065}{0.71}=\mathbf{0.0915} \]

The managerial reading: nearly three-quarters of collections come from personal calls — but that is mostly because personal calls are used most often (70%), not because they are much more effective (0.75 vs 0.65 for a letter). Do not confuse the posterior with the effectiveness; letters collect 65% of the time at a fraction of the cost.

4-43"Which most likely?"Levin Ex. 4-43 · p. 194

Q. P(choose New Delhi) = 0.40, Mumbai 0.35, Chennai 0.25. P(favourable ruling) = 0.45, 0.60, 0.35 respectively. Given that the group received a favourable ruling, which city did it most likely choose?

Why: "which most likely, given the evidence" is a request for the largest posterior — and the answer flips the prior ordering, which is the point.

City	Prior	P(favourable \| city)	Joint	Posterior
New Delhi	0.40	0.45	0.1800	0.3770
Mumbai	0.35	0.60	0.2100	0.4398
Chennai	0.25	0.35	0.0875	0.1832
Total	1.00	—	0.4775	1.0000 ✓

AnswerMumbai, with posterior probability 0.4398.

The insight that earns the mark: New Delhi was the most likely choice beforehand (prior 0.40 vs 0.35). But a favourable ruling is much more likely in Mumbai (0.60 vs 0.45), and that evidence is enough to overturn the ordering. This is exactly what Bayes does — evidence revises belief.

4-44Total probability + posteriorLevin Ex. 4-44 · p. 194

Q. P(dry) = 0.2, P(moist) = 0.45, P(wet) = 0.35. P(thunderstorm) given each is 0.3, 0.6, 0.8. What is the probability of a thunderstorm? Given the picnic was cancelled, what is the probability conditions were moist?

Why: the two-part shape the exam uses — first the law of total probability, then one posterior. The cancellation is the thunderstorm, since that is the only thing that cancels the picnic.

Why we do itWhat to write

Build the joints.

dry: \(0.20(0.3)=0.060\)
moist: \(0.45(0.6)=0.270\)
wet: \(0.35(0.8)=0.280\)

Total them — this answers the first question directly.

\[ P(\text{thunderstorm})=0.060+0.270+0.280=\mathbf{0.61} \]

The posterior for moist.

\[ P(\text{moist}\mid\text{cancelled})=\frac{0.270}{0.61}=\mathbf{0.4426} \] Full set: dry 0.0984, moist 0.4426, wet 0.4590 — sum 1.0000 ✓

The reading: before knowing anything, moist was the most likely condition (0.45). After the cancellation, wet becomes marginally more likely (0.459 vs 0.443) — because a thunderstorm is more probable under wet conditions. The evidence shifts belief toward the cause that explains it best.

4-45Joints given, priors hiddenLevin Ex. 4-45 · p. 194

Q. Accidents are fire, mechanical failure, or human error, never together. P(leak) given each is 0.20, 0.50, 0.10. The probability of a fire and a leak together is 0.0010; mechanical failure and leak 0.0015; human error and leak 0.0012. Find (a) P(fire), P(mech), P(human); (b) P(each | leak); (c) P(leak).

Why: the table is handed to you middle-out — you get the joints and the likelihoods, and must work backwards to the priors. Recognising which column you have been given is the whole difficulty.

Why we do itWhat to write

(a) Since joint = prior × likelihood, the prior is joint ÷ likelihood.

\[ P(\text{fire})=\frac{0.0010}{0.20}=\mathbf{0.005} \] \[ P(\text{mech})=\frac{0.0015}{0.50}=\mathbf{0.003} \] \[ P(\text{human})=\frac{0.0012}{0.10}=\mathbf{0.012} \]

\[ P(\text{leak})=0.0010+0.0015+0.0012=\mathbf{0.0037} \]

(b) Each joint over that total.

\[ P(\text{fire}|\text{leak})=\tfrac{0.0010}{0.0037}=\mathbf{0.2703} \] \[ P(\text{mech}|\text{leak})=\tfrac{0.0015}{0.0037}=\mathbf{0.4054} \] \[ P(\text{human}|\text{leak})=\tfrac{0.0012}{0.0037}=\mathbf{0.3243} \] Sum \(=1.0000\) ✓

The safety insight: human error is the most likely accident (prior 0.012, four times more likely than fire) but it leaks only 10% of the time. Mechanical failure is the rarest accident yet the most likely cause of a leak (0.405). So the plant should spend on mechanical reliability to prevent leaks, and on training to prevent accidents — two different priorities, and only Bayes separates them.

4-46Derive the likelihoodLevin Ex. 4-46 · p. 195

Q. 40% of games are on artificial turf. Knee-injury chances are 50% higher on turf than on grass, and P(injury | turf) = 0.42. Find (a) P(a player incurs a knee injury), (b) P(the injury was incurred on grass | injured).

Why: one likelihood is missing and must be derived from "50% higher". If turf is 1.5 times grass, then grass = turf ÷ 1.5 — divide, do not subtract 50%.

Why we do itWhat to write

Derive the grass likelihood. "Turf is 50% higher than grass" means \(0.42 = 1.5 \times P(\text{injury}|\text{grass})\).

\[ P(\text{injury}|\text{grass})=\frac{0.42}{1.5}=\mathbf{0.28} \] Not \(0.42-0.21=0.21\) — that would be "50% of", not "50% higher than".

Build the joints. Turf 40%, grass 60%.

turf: \(0.40(0.42)=0.168\)
grass: \(0.60(0.28)=0.168\)

(a) Total.

\[ P(\text{injury})=0.168+0.168=\mathbf{0.336} \]

(b) Posterior for grass.

\[ P(\text{grass}|\text{injury})=\frac{0.168}{0.336}=\mathbf{0.50} \]

The striking result: exactly half of knee injuries happen on grass, even though grass is much safer per game — because 60% of games are played on it. A safer surface used more often produces just as many injuries. That is the managerial sentence the question wants.

4-47From raw countsLevin Ex. 4-47 · p. 195

Q. Foot injuries by position — offensive line 45 players / 32 injured; defensive line 56 / 38; offensive backfield 24 / 11; defensive backfield 20 / 9. Given a player incurred a foot injury, what is the probability he plays in each position?

Why: Bayes with the joints already counted for you. The injured counts are the joints (up to a common denominator), so the posteriors are just each count over the total injured.

Why we do itWhat to write

Total the injured — this is the denominator, not the 145 players.

\[ 32+38+11+9=\mathbf{90 \text{ injured}} \]

Each position's share of the injuries.

\[ \text{(a) offensive line } \tfrac{32}{90}=\mathbf{0.3556} \] \[ \text{(b) defensive line } \tfrac{38}{90}=\mathbf{0.4222} \] \[ \text{(c) offensive backfield } \tfrac{11}{90}=\mathbf{0.1222} \] \[ \text{(d) defensive backfield } \tfrac{9}{90}=\mathbf{0.1000} \] Sum \(=1.0000\) ✓

Do not stop there — the interesting comparison is the injury rate: offensive line 32/45 = 71%, defensive line 38/56 = 68%, offensive backfield 11/24 = 46%, defensive backfield 9/20 = 45%. Linemen are injured at a far higher rate; they also dominate the posterior partly because there are more of them. Posterior share and risk rate are different questions, and a good answer distinguishes them.

4-48Sequential evidenceLevin Ex. 4-48 · p. 195

Q. A film is a hit (0.60), moderate (0.25) or flop (0.15). Rating probabilities — hit: 0.60 / 0.30 / 0.10 for scores 7+, 4–6, ≤3; moderate: 0.30 / 0.45 / 0.25; flop: 0.15 / 0.35 / 0.50. (a) If the first screening scores 6, what is P(hit)? (b) If the first scores 6 and the second scores 2, what is P(flop)?

Why: the hardest shape — two pieces of evidence. Because the screenings are independent, the combined likelihood under each cause is simply the product of the two individual likelihoods.

Why we do itWhat to write

(a) A score of 6 falls in the 4–6 band. Build the table with those likelihoods.

hit: \(0.60(0.30)=0.1800\)
moderate: \(0.25(0.45)=0.1125\)
flop: \(0.15(0.35)=0.0525\)
Total \(=\mathbf{0.3450}\)

Posterior for hit.

\[ P(\text{hit}\mid 6)=\frac{0.1800}{0.3450}=\mathbf{0.5217} \] Down from a prior of 0.60 — a middling score is mild evidence against a hit.

(b) Two independent screenings: multiply the two likelihoods for each cause. Score 6 → 4–6 band; score 2 → ≤3 band.

hit: \(0.30\times0.10=0.030\)
moderate: \(0.45\times0.25=0.1125\)
flop: \(0.35\times0.50=0.175\)

Now weight by the priors and total.

hit: \(0.60(0.030)=0.01800\)
moderate: \(0.25(0.1125)=0.02813\)
flop: \(0.15(0.175)=0.02625\)
Total \(=\mathbf{0.07238}\)

Posterior for flop.

\[ P(\text{flop}\mid 6\text{ then }2)=\frac{0.02625}{0.07238}=\mathbf{0.3627} \] Full set: hit 0.2487, moderate 0.3886, flop 0.3627 — sum 1.0000 ✓

Watch the belief move: flop started at a prior of 0.15. One middling score barely changed it (0.152); a second, bad score more than doubled it to 0.363, while "hit" collapsed from 0.60 to 0.25. Evidence accumulates multiplicatively — that is why two weak signals can overturn a strong prior.

SC 4-11Standard tableLevin SC 4-11 · p. 196

Q. Priors A 0.35, B 0.45, C 0.20; likelihoods of X given each 0.80, 0.65, 0.30. Find the three posteriors.

Event	P(Event)	P(X \| Event)	Joint	P(Event \| X)
A	0.35	0.80	0.2800	0.4427
B	0.45	0.65	0.2925	0.4625
C	0.20	0.30	0.0600	0.0949
Total	1.00	—	0.6325	1.0001 ✓

Matches Levin's printed answers (0.4427, 0.4625, 0.0949) exactly. This table layout is what to reproduce in the exam — five columns, a ruled total row, and the check that the posteriors sum to 1.

SC 4-12Compound likelihoodsLevin SC 4-12 · p. 196

Q. Heart-attack risk factors A, B, or both, with priors 0.25, 0.25, 0.50. The likelihood of a heart attack is a product of factor effects: A gives (0.8)(0.65), B gives (0.8)(0.80), and A&B gives (0.8)(0.65)(0.80). Find P(A&B | heart attack).

Why: the likelihoods themselves are products — but the method is unchanged. Compute each likelihood first, then run the standard table.

Event	P(Event)	P(H \| Event)	Joint	P(Event \| H)
A	0.25	(0.8)(0.65) = 0.520	0.1300	0.2610
B	0.25	(0.8)(0.80) = 0.640	0.1600	0.3213
A & B	0.50	(0.8)(0.65)(0.80) = 0.416	0.2080	0.4177
Total	1.00	—	0.4980	1.0000 ✓

Answer\(P(A\&B \mid \text{heart attack}) = \mathbf{0.4177}\) — matching Levin's printed answer exactly.

Note the subtlety: A&B has the lowest likelihood of the three (0.416) yet the highest posterior, because its prior (0.50) is twice the others. A common cause with a modest likelihood beats a rare cause with a high one — the base-rate effect, which is the single most examinable idea in Bayes.

THE 90-SECOND VERSION — WHAT TO CARRY INTO THE HALL

OR P(A or B) = P(A) + P(B) − P(AB) subtract the overlap mutually exclusive → P(AB) = 0 AND independent: P(AB) = P(A) × P(B) dependent: P(AB) = P(A|B) × P(B) GIVEN independent: P(B|A) = P(B) dependent: P(B|A) = P(AB)/P(A) AT LEAST ONE P = 1 − P(none) = 1 − ∏(1 − pᵢ) never sum the cases INDEPENDENCE TEST P(AB) = P(A)P(B) ⟺ P(B|A) = P(B) BAYES — Q5 (a) — build the TABLE, not the formula: cause | prior | likelihood | joint = prior×lik | posterior = joint/TOTAL TOTAL = P(evidence) 1. priors must sum to 1 — recover a missing one 2. likelihood = P(evidence | that cause) 3. multiply across, total the joint column 4. divide each joint by the total 5. CHECK posteriors sum to 1 6. SAY WHAT IT MEANS — that sentence is a mark CONTINGENCY TABLE joint → divide by GRAND total conditional → divide by the total of the GIVEN category

◆ Where this sits in the rest of the site

Concepts: Unit 03 notes. Q5 (a) is worked in both its versions at Solved Papers, and your class's heart-attack and rare-disease examples are in Unit 03, Concept 4.

Unit 3 feeds Unit 4 directly: the binomial is repeated independent trials, and the Poisson approximation rests on the same independence assumption you have been checking all through Topic 3.