Partitioning the sample space, and totalling a probability across the parts

What to assume they know

• Conditional probability, its definition and its side condition
• The product form of the multiplication rule from §13.3
• Disjoint events, and that the probability of a union of disjoint events is a sum
• An event and its complement, and that the two exhaust the space
• Intersection distributing across a union of several sets
• Sigma notation for a finite sum, from Class XI

What they should be able to do

• State the question §13.5 opens with, and say which direction of conditioning is already available and which is not
• List the three demands a cut of the sample space has to meet
• Say, for each demand separately, what goes wrong in the proof without it
• Verify that an event and its complement meet all three
• Read a four-piece cut off two overlapping events
• Give two different cuts of the same space and explain why non-uniqueness is harmless
• State the totalling result, in longhand and in sigma notation
• Reproduce its four-move proof, naming the step that uses each demand
• Apply the result to a two-piece cut where one conditional and its complement's conditional are both given
• Recognise the same total appearing as the denominator of everything in the next topic

Where it usually goes wrong

• "A partition is just any collection of events." It is three conditions at once, and dropping any one breaks a specific line of the proof. Name the line each time.
• "The pieces have to be equally likely." They almost never are. Example 15's two pieces are nought point six five and nought point three five, and Fig 13.4 draws its cells at visibly different sizes for exactly this reason.
• "The pieces have to be the obvious physical categories." The chapter says twice that a space can be cut more than one way. Which cut you choose is a modelling decision, and a good one makes the conditionals easy to write down.
• "Totalling means adding the conditionals." It means adding the weighted conditionals. Adding the bare conditionals in Example 15 would give one point one two, which is not a probability at all — show that failure once.
• "The answer can come out above the biggest conditional." It cannot. It is a weighted average and always lands between the smallest and the largest of them. This is the fastest sanity check available and the chapter never mentions it.
• "This is a new rule to memorise." It is the multiplication rule applied once per piece and then added up. If the proof is shown, nothing needs memorising.
• "The event has to lie inside one of the pieces." It generally straddles several — that is the whole point, and Fig 13.4 draws it straddling.
• "An event with probability zero can be a piece as long as it fits." The third demand forbids it, because that piece's conditional would be undefined and its term unwritable.

Questions to check understanding

• State the three demands on a partition and give one family failing each demand in turn
• Verify that a given family of events is or is not a partition
• Read a four-piece partition off two overlapping events
• State the totalling result in sigma notation and expand it for three pieces
• Prove the result, naming the demand each step uses
• Given a two-branch situation with both conditionals supplied, compute the total — the form of Example 15
• Check an answer by confirming it lies between the conditionals it was built from
• Work a two-stage draw where the second stage's contents depend on the first — the form of Exercise 13.3 Q1

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 opening setup (§13.5, Part II p. 423). Two bags with stated numbers of white and red balls, one bag chosen at random, one ball drawn. The chapter says outright which probabilities are easy — the bag, and the colour once the bag is named — and then asks for the one that is not: the bag, once the colour is known. That single sentence is the motivation for the whole of module three. The chapter never returns to these bags. Example 16 on Part II p. 426 uses a different pair of bags with different contents. Verified, if the explanation chooses to close the loop: with the opening numbers, a white ball came from the second bag with probability ten nineteenths. That is worked out here, not the book's, and it must be labelled so.
• The three demands (§13.5.1, Part II p. 423). Listed as (a), (b) and (c): no two pieces share an outcome; the pieces together make up the whole space; every piece has positive probability. The chapter then restates all three in one sentence of prose on the next page, using three ordinary words. Put the symbolic list and the prose sentence together — students who bounce off one will land on the other.
• What each demand pays for. This is an added analysis; the chapter offers no such remark. No overlap is what turns the probability of the union into a sum, in the fourth line of the proof. Covering everything is what makes the first line true, since the event has to be recoverable as its own intersection with the whole space. Positive probability is what lets the last line write each term as a weight times a conditional, because the conditional is undefined otherwise. Show the proof once, then run it again with each demand struck out in turn and the corresponding line failing.
• The smallest partition (Part II p. 424). Any nonempty event and its complement. The chapter checks both of the set conditions explicitly and is silent about the third. Verified: the third holds exactly when neither of the two is a null event — the chapter's word nonempty is doing that job for the first of them and nothing is said about the second. Worth thirty seconds, because every two-branch problem in the rest of the chapter uses this partition and only this one.
• The four-piece partition (Part II p. 424). The chapter reads a four-member partition off two overlapping events by pointing back at Fig 13.3, which was drawn three pages earlier for a different argument. Read off that figure: the four regions are labelled with arrows — outside both, first only, both, second only — and the labels are exactly the four pieces named in the §13.5.1 sentence. The figure's shading, however, fills only the two crescents and leaves the overlap white, which is a leftover from the argument it was drawn for. Redraw with all four regions distinct.
• Non-uniqueness (Part II p. 424). Two flat sentences saying a space can be cut more than one way. No example is given. The same space cut two ways, with the same event totalled across both cuts, landing on the same number. That makes non-uniqueness a feature rather than a worry.
• The totalling result and its proof (§13.5.2, Part II pp. 424–425). Statement first in longhand and then in sigma notation. The proof is four moves: rewrite the event as its intersection with the whole space; distribute across the pieces; add, using disjointness inherited from the cut; and replace each term by a weight times a conditional. Note that the distributive step here is done silently, whereas the same step in §13.2.1 on Part II p. 409 carries a bracketed justification — and that justification names the two operations the wrong way round. Prefer this one.
• Fig 13.4 (Part II p. 424). Read off the printed page: a rectangle labelled as the sample space, cut by curved lines into cells labelled for the first, second and third pieces, an ellipsis, and the last piece; a shaded ellipse labelled for the event lies across several cells. The dividing lines are drawn only outside the ellipse, so the event reads as one region overlapping many pieces. The cells are visibly unequal in size, which is the right picture: the pieces of a partition are not required to be equally likely.
• Example 15 (Part II p. 425), a construction job. The chance of a strike is given, and the chance of finishing on time is given separately for the struck and the unstruck case. Verified: the complement probability is nought point three five, the two products are nought point two nought eight and nought point two eight, and the total is nought point four eight eight. Note what makes this a good closing example — the answer sits between the two conditionals it was built from, which is what a weighted average always does, and that is a check a student can apply without redoing the arithmetic.
• Exercise 13.3 Q1 (Part II p. 431). An urn with equal numbers of two colours; a ball is drawn, put back, and two more of the colour drawn are added; then a second ball is drawn. This is the only item in Exercise 13.3 that stops at the total and does not go on to reverse anything, which makes it the natural practice item for this topic. Verified: the second draw comes from a container of twelve, holding seven of the drawn colour and five of the other, so the total is one half of seven twelfths plus one half of five twelfths, which is one half — the same as the first draw. Show that coincidence; it is memorable and it is true for a reason worth a sentence.
• The Summary's version (Part II p. 437). The fifth bullet restates the totalling result with its hypotheses intact. It is one of the two places in the Summary where nothing was lost relative to the body.

Figures to have open

• Two bags for section 1, drawn once and reused in section 2 with the arrow direction flipped, so the reversal is a change to one drawing rather than a new one. The chapter draws no figure for its opening setup.
• A single space rectangle for sections 3 to 7, cut and re-cut in place. Every cut must be drawn into the same rectangle at the same size, or non-uniqueness reads as three unrelated pictures.
• A redrawn four-region diagram for section 6 with all four pieces distinctly filled. Do not reproduce Fig 13.3's shading, which leaves the overlap white for reasons belonging to a different proof.
• A partition-and-event figure for sections 8 and 9 in the spirit of Fig 13.4: unequal cells, one event straddling them, dividing lines suppressed under the event. Build with the repo's Shells or Network component depending on whether the emphasis is nesting or routes.
• A two-branch tree for section 11 built with Network, carrying conditional chances on its edges. It must be the same drawing the next topic reverses, so fix the layout here.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 13 "Probability", §13.5 Bayes' Theorem, the opening two-bag setup, Part II p. 423
• §13.5.1 Partition of a sample space: the three conditions, Part II p. 423; the prose restatement, the complement example, the four-piece partition and the non-uniqueness sentences, Part II p. 424
• §13.5.2 Theorem of total probability: the statement and the proof, Part II pp. 424–425; Fig 13.4, Part II p. 424
• Example 15, Part II p. 425
• Exercise 13.3, question 1, Part II p. 431
• Summary, fifth bullet, Part II p. 437