Working back from an observed effect to the likelihood of each cause

What to assume they know

• The definition of a conditional probability and its side condition
• The product form of the multiplication rule from §13.3
• Partitions and the total across a partition, from the previous topic
• Complement of an event, and the two-piece partition it gives
• Sigma notation for a finite sum
• Converting percentages to probabilities, and multiplying small decimals

What they should be able to do

• State the reversal formula with all of its hypotheses
• Derive it in three lines, naming the earlier result each line uses
• Identify the denominator as the total from the previous topic and say why it is never anything else
• Name the three roles the chapter's terminology gives — the pieces, the probability before, the probability after
• Set a word problem up by naming the partition first and the observed event second, in that order
• Work a two-branch reversal from stated bag contents
• Explain why a highly accurate test on a rare condition still gives a small answer, in terms of the two products in the denominator
• Work a three-branch reversal of the manufacturing kind
• Handle a branch whose conditional is zero, and say what it contributes
• Recognise the same computation inside a puzzle whose wording hides it

Where it usually goes wrong

• "The two conditionals are interchangeable." The chance of a positive result given the condition and the chance of the condition given a positive result are different numbers, and in Example 18 they are nought point nine and nought point nought eight three. This confusion is the reason the topic exists.
• "A ninety per cent accurate test means a positive result is ninety per cent likely to be right." It does not, and the gap is the whole content of Example 18. Show the two branch populations before showing any fraction.
• "The denominator is the probability of the piece." It is the total of every piece's contribution. Building it from the wrong set of pieces is the single commonest structural error, and it is invisible in a written answer because the fraction still looks like a fraction.
• "The pieces have to be equally likely for the formula to work." They never have to be. Example 16 happens to have equal weights and Examples 19, 20 and 21 do not; do Example 16 first and then break the symmetry deliberately.
• "A branch whose conditional is zero has to be dropped from the partition." It stays in the partition, contributes nothing to the total, and comes out with answer zero. Example 20 has exactly such a branch.
• "Once the answer is computed, the before-probabilities are irrelevant." They are half the formula. In Example 21 the before-probabilities are one sixth and five sixths and they are what pull the answer below a half.
• "The gold-coin puzzle is about coins." It is about which box was chosen, and the chapter's own solution says so in one line that a student will read past. Slow down there.
• "Every question printed after this section is a reversal question." Exercise 13.3 Q1 stops at the total and reverses nothing; it belongs to the previous topic.

Questions to check understanding

• State the reversal formula with all its hypotheses and derive it in three lines
• Given a word problem, name the partition and the observed event before writing any number
• Work a two-branch reversal from stated container contents — the form of Example 16
• Work a three-branch or four-branch reversal from stated shares and rates — the form of Examples 19 and 20
• Given a test's detection rate, its false-flag rate and how common the condition is, compute the chance a flagged person has it, and explain the size of the answer in terms of the two products
• Identify which branch of a partition can be discarded from the arithmetic because its conditional is zero, and say why it stays in the partition
• Recast a puzzle about what is left in a container as a question about which container was chosen — the form of Example 17

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The statement (Part II p. 425). Printed as a named result rather than a numbered theorem — the chapter numbers no theorems at all. It demands a partition into nonempty pieces and an observed event of positive probability, and it returns one answer per piece.
• The proof, three lines (Part II p. 425). Write the wanted conditional from the definition; replace the numerator using the product rule; replace the denominator using the total across the partition. Each line is annotated on the page with the result it invokes. This is the cleanest short proof in the chapter.
• The denominator point. An added emphasis. The bottom of the fraction is the whole of the previous topic, unchanged, and the top is one of its own terms. Draw the total once, then lift one term out of it and set it above the whole. Students who see that never again build a denominator out of the wrong pieces, which is the commonest structural error in this material.
• The Remark (Part II p. 426). Three pieces of vocabulary — a collective name for the pieces, a name for their probabilities before, and a name for their probabilities after — plus a sentence saying exactly one of the pieces actually happens, which is the partition property restated in words. The forename attached to the result on Part II p. 423 is wrong and the article in front of one of these three terms is wrong; see Notes for both.
• Example 16 (Part II p. 426), two bags of red and black balls, a red ball drawn, the second bag wanted. Verified: with equal chance of either bag, the two conditionals are three sevenths and five elevenths, and the answer is thirty-five over sixty-eight. Note the bag choice cancels — both weights are one half — which makes this the right first example, because the arithmetic is visibly just the two conditionals compared.
• Example 17 (Part II pp. 426–427), three boxes each holding two coins: two gold, two silver, one of each. A coin is drawn and is gold; what is wanted is the chance that its partner in the same box is gold too. Verified: the three conditionals are one, nought and one half, and the answer is two thirds. The step that does the real work is a single printed sentence identifying the wanted event with the event that the all-gold box was chosen. The chapter states that identification and does not argue for it. Argue for it: if the other coin is gold, the box had two gold coins; and conversely. Everything else is arithmetic.
• Example 18 (Part II pp. 427–428), a test for a condition. Nine in ten cases detected; one in a hundred healthy people wrongly flagged; one in a thousand of the population affected. Verified: the two products are nought point nought nought nought nine and nought point nought nought nine nine nine, so the answer is ninety over one thousand and eighty-nine, about nought point nought eight three. This is the most important number in the chapter and the chapter does not comment on it. The reason is entirely visible in the denominator: the healthy branch is nine hundred and ninety-nine times more populated than the other, so even a one-in-a-hundred error rate on that branch produces eleven times as many flags as the true cases do. Put the two products side by side at full size before dividing.
• Example 19 (Part II pp. 428–429), three machines making bolts with different output shares and different defect rates. Verified: the three products are nought point nought one two five, nought point nought one four and nought point nought nought eight, totalling nought point nought three four five, so the answer is twenty-eight over sixty-nine. Note the chapter names the machines with letters and then names the events with a different letter carrying subscripts — worth mirroring, because keeping the physical object and the event visually distinct is half the battle in these problems.
• Example 20 (Part II p. 429), a doctor arriving late by one of four modes of transport. Verified: the four products are three fortieths, one fifteenth, one hundred and twentieth and zero, totalling eighteen over one hundred and twenty, so the answer is one half. The fourth branch has conditional zero, and this is the example to use for section 11: a piece of the partition must have positive probability, but nothing stops its conditional being zero, and such a branch contributes nothing to the denominator and can never be the answer.
• Example 21 (Part II p. 430), a man who speaks truthfully three times in four reporting a six on a die. Verified: the two products are one eighth and five twenty-fourths, totalling one third, so the answer is three eighths — less than a half, which surprises people, because the report is more often true than false and yet a six is rare. It is Example 18's lesson in a form that fits on one line, and it is the right closing beat.
• Exercise 13.3, questions 2 to 14 (Part II pp. 431–433). All thirteen reversal items were worked. Verified: Q2 two thirds; Q3 nine thirteenths; Q4 twelve thirteenths; Q5 twenty-two over one hundred and thirty-three, which is another rare-condition case and about nought point one seven; Q6 four ninths; Q7 one over fifty-two; Q8 one quarter; Q9 two ninths; Q10 eight elevenths; Q11 five over thirty-four; Q12 eleven over fifty; Q13 four fifths, which is option (A); Q14 the option stating that conditioning on a containing event cannot lower the probability, which is option (C). Two of these need machinery from outside the chapter and are flagged in Notes.
• The Summary's version (Part II p. 437). The last bullet restates the formula with its hypotheses intact and is faithful to the body.

Figures to have open

• One tree drawn in section 1 and reused, unchanged in layout, in sections 4, 11 and 12. If the tree is redrawn each time the reversal reads as four separate tricks. Build with the repo's Network component and fix the geometry once.
• A population figure for section 9: a thousand figures, one branch of them marked as affected, with the true detections and the wrong flags each picked out. This is an added device; the chapter draws nothing for Example 18. It must be drawn before the fraction appears, or the fraction is doing the teaching and the picture is decoration.
• A six-coin figure for section 8, arranged as three boxes of two, with the drawn coin shown as gold and the three candidate boxes narrowed to two. Not in the book.
• A three-row table for section 10 with output share, defect rate and product as the columns. Build with DataTable, and give all three rows one shared fitSize.
• No new figure for sections 2 and 3. They are notation and proof, and Statement and Steps carry them.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 13 "Probability", the named result and its proof, Part II p. 425
• The Remark on terminology and the alternative name for the formula, Part II p. 426
• Examples 16 and 17, Part II pp. 426–427
• Example 18, Part II pp. 427–428
• Examples 19 and 20, Part II pp. 428–429
• Example 21, Part II p. 430
• Exercise 13.3, questions 2 to 14, Part II pp. 431–433
• §13.5's opening paragraph, Part II p. 423; Summary, last bullet, Part II p. 437