Adding two probabilities double-counts the overlap, so the overlap comes back off

What to assume they know

• Three conditions on a function, in place of a recipe for counting — the three axioms, and that the third one requires a non-overlapping pair
• Assuming the outcomes are equally likely recovers the counting rule — an event's probability is the total over its outcomes
• "Or", "and" and "not" are the three set operations under new names — union, intersection and difference as operations on events
• Events that cannot both happen, and events that between them must — what it means for events to be unable to occur together
• That a set equals the union of the parts inside and outside another set, and the distributive law relating intersection over union
• Reading a Venn diagram with two overlapping regions inside a rectangle

What they should be able to do

• Show by a worked case that totalling two probabilities overshoots the union, and identify the overshoot as the overlap
• Cut a union of two events into three parts no two of which share an outcome, and name each part
• Prove the two-event rule by applying the additivity axiom to those three parts
• Reproduce the chapter's shorter proof, which subtracts one equation from another
• Recover the additivity axiom as the case where the overlap is empty
• Apply the rule where the intersection is given and the union is wanted, and in reverse
• Decide whether stated values for two events and their overlap can hold together
• Extend the rule to three events by applying the two-event rule twice

Where it usually goes wrong

• "Probabilities of two events always add." They add when the events cannot both occur, and not otherwise. The chapter's opening pair overshoots by a quarter, which is a large error on a probability of a half.
• "Subtracting the overlap is an adjustment someone noticed empirically." It falls out of cutting the union into three non-overlapping parts, which is the only shape the axiom accepts. The chapter gives two proofs rather than one.
• "The overlap is removed twice, since it was in both events." It was counted twice, so it must be removed once, leaving it counted once. Removing it twice would lose it entirely.
• "If two probabilities total more than 1, something is wrong." Nothing is. Exercise 14.2 Q19 has 0.8 and 0.7 for two examinations. Only a single event's value is capped at 1, and the union's value is what the rule brings back under the cap.
• "Any three numbers can serve as the two events and their overlap." The overlap sits inside both, so it cannot exceed either, and Q12's first set breaks that outright.
• "Cannot occur together is the same as unrelated." They are different ideas, and this chapter contains only the first. Neither independence nor conditional probability is defined anywhere in it — checked against the page images of all of pp. 289–313. The word that looks like independence on p. 313 describes two mathematicians working separately, not events.
• "Exactly one of them qualifies means one minus both qualify." That is at least one failing, which is Example 7's second part and comes to 0.98. Exactly one is 0.11, and the two are answers to different questions.
• "Fig 14.1 shades the overlap because the overlap is the answer." The printed figure shades the two crescents and leaves the lens white. Redrawing it with the lens shaded inverts the reading of the picture.
• "The three-event rule needs its own axiom." It is derived in Example 11 from the two-event rule alone, using a set identity and nothing more.

Questions to check understanding

• Given two of the four quantities and the rule, find a third — Exercise 14.2 Q13 sets three such rows
• Exercise 14.2 Q14 gives 3/5 and 1/5 for a pair that cannot both occur and asks for the union
• Exercise 14.2 Q15 gives 1/4, 1/2 and an overlap of 1/8, asking for the union and then for the probability that neither occurs
• Exercise 14.2 Q17 gives 0.42, 0.48 and an overlap of 0.16
• Exercise 14.2 Q18 puts 40 per cent of a class on one subject, 30 per cent on another and 10 per cent on both, asking for either
• Exercise 14.2 Q19 gives 0.8 and 0.7 for two examinations with 0.95 for passing at least one, and asks for both — the rule run backwards
• Exercise 14.2 Q20 gives 0.5 for passing both and 0.1 for passing neither, with 0.75 for one of the two subjects, and asks for the other
• Exercise 14.2 Q21 has 60 students, 30 on one activity, 32 on another and 24 on both, asking for either, for neither, and for the second without the first
• Miscellaneous Exercise Q7 gives 0.54, 0.69 and an overlap of 0.35 and asks for four combinations
• Miscellaneous Exercise Q8 lists five named people with sex and age, and asks for male or over 35 — a case where the two descriptions overlap and the overlap has to be spotted from the table
• Decide whether stated values can hold together, and justify by the bound on the overlap

Examples worth working on the board

Values marked verified are worked out here against the printed page.

• The motivating pair (§14.2.3, p. 299). Three tossed coins, all eight outcomes equally weighted at 1/8. The chapter supplies two events as bare lists, with no description attached to either. Written one per line so they can be compared member by member:
  • the first event holds HHT, HTH, THH
  • the second holds HTH, THH, HHH

Verified: each holds three outcomes, so each is worth 3/8, and the two total 6/8. Two of the three members are common to both.

• The union (§14.2.3, p. 299). A ∪ B = {HHT, HTH, THH, HHH}. Verified: four outcomes, so 4/8, which is 1/2 — and 4/8 is not 6/8. The chapter states the inequality plainly rather than glossing it.
• The overlap (§14.2.3, p. 300). A ∩ B = {HTH, THH}. Verified: two outcomes, so 2/8, which is 1/4. The excess of the total over the union is 6/8 less 4/8, which is 2/8 — exactly the overlap. That equality is the whole discovery, and running it in eighths makes it unmissable.
• The three-part cut (§14.2.3, p. 300). The union divides into what lies in the first event only, what lies in both, and what lies in the second only. Verified on the same example, and not printed: those parts are {HHT}, {HTH, THH} and {HHH}, of sizes 1, 2 and 1, totalling the four outcomes of the union. No outcome appears in two parts, so the axiom may be applied across them.
• The proof (§14.2.3, p. 300). Totalling over the first event covers the first part and the shared part; totalling over the second covers the shared part and the third. Verified: adding those two totals therefore covers the union once and the shared part one extra time, so removing the shared part's value once leaves the union exactly.
• The shorter proof (§14.2.3, pp. 300–301). The union splits as the first event together with what is in the second but not the first, and those two share nothing. The second event splits as the shared part together with the same leftover, and those two share nothing either. Verified: both splits give an equation from the axiom directly, and subtracting one from the other cancels the leftover and rearranges into the rule.
• Fig 14.1 (p. 301). A rectangle labelled S carrying two overlapping circles, labelled A on the left with an arrow and B on the right with an arrow. Three regions are named inside: the left crescent, the central lens and the right crescent. In the printed figure the two crescents are shaded pale blue and the central lens is left white — checked on the printed page, because the shading is what tells a reader the crescents are the parts being added and the lens is the part in dispute.
• The special case (§14.2.3, p. 301). Verified: when the two events share nothing the overlap is the impossible event, whose value the axioms already force to 0, so the subtraction removes nothing and the rule reads as the axiom itself.
• Example 7 — two candidates (pp. 304–305). Anil's event is given 0.05, Ashima's 0.10, and both qualifying 0.02. Verified: at least one qualifies is 0.05 plus 0.10 less 0.02, so 0.13; neither qualifies is what remains, so 0.87; at least one fails is what remains after both qualifying, so 0.98; and exactly one qualifies is 0.03 plus 0.08, so 0.11. Verified, and not printed: 0.11 is also 0.13 less 0.02, since the union minus the overlap is precisely the outcomes belonging to one event and not the other. Showing both routes to 0.11 is a free check.
• Example 11 — three events (pp. 309–310). Verified: grouping the second and third events into a single event lets the two-event rule be applied once, then again inside; distributing the intersection over that union produces two intersections whose own overlap is the triple intersection. The result carries the three single values, less the three pairwise overlaps, plus the triple one. No fourth axiom is introduced — the whole extension is the two-event rule used twice.
• Consistency (Exercise 14.2 Q12, p. 307). Two sets of figures are offered: 0.5, 0.7 with an overlap of 0.6; and 0.5, 0.4 with a union of 0.8. Verified: the overlap sits inside each event, so it cannot be worth more than either — 0.6 exceeds 0.5, so the first set cannot hold together. The second gives an overlap of 0.1, which sits below both 0.5 and 0.4 and is not negative, so it can.
• The blank-filling drill (Exercise 14.2 Q13, p. 307). Three rows, each missing one of the four quantities: 1/3, 1/5 and an overlap of 1/15; 0.35, a missing value, an overlap of 0.25 and a union of 0.6; and 0.5, 0.35, a missing overlap and a union of 0.7. Verified: the answers are a union of 7/15, a second event worth 0.5, and an overlap of 0.15.

Figures to have open

• Fig 14.1 (p. 301) — the chapter's only numbered figure, and this topic's central asset. A rectangle labelled S, two overlapping circles labelled A and B by arrows from outside, and the three regions named. Redraw it as a schematic rather than reproducing the printed art, but keep the shading as printed: both crescents shaded, the lens unshaded.
• An eight-cell strip for the three-coin outcome list, with cells liftable into two overlapping groups. An added device.
• An exploded version of Fig 14.1 with the three regions pulled apart, for the proof. Not in the book.
• A two-by-two panel for Example 7 covering both qualifying, neither, and each one alone. Not in the book; the chapter works Example 7 in running text with no picture, verified against the page images of pp. 304–305.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 14 "Probability", §14.2.3 "Probability of the event 'A or B'", pp. 299–301, comprising the motivating pair, the three-part proof, Fig 14.1 on p. 301, the shorter proof and the no-overlap case
• Example 7, pp. 304–305
• Miscellaneous Example 11, pp. 309–310, for the three-event extension
• Exercise 14.2 Q12 to Q21, pp. 307–308; Miscellaneous Exercise Q7 and Q8, p. 311
• The Summary, p. 312, which states the rule in both the wording form and the set form and then states the no-overlap case separately