Knowing one probability hands you its opposite for free

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What to assume they know

Three conditions on a function, in place of a recipe for counting — the three axioms, in particular the value of the sure event and additivity across a non-overlapping pair
Adding two probabilities double-counts the overlap, so the overlap comes back off — the two-event rule, which this topic is chained with
"Or", "and" and "not" are the three set operations under new names — the complement of an event, and the rule exchanging a complemented union for an intersection of complements
Events that cannot both happen, and events that between them must — events that cannot both occur and that between them cover everything
Selections from the counting chapter earlier in this book, for the exercises where the omitted outcomes must be counted rather than listed

What they should be able to do

Prove that an event and its complement have probabilities totalling 1, naming the two axioms used
Write the complement of a stated event as an explicit list, and check the two sizes total the size of the sample space
Explain why the complement need not be describable by any single shared property of its members
Choose, given a question, whether to compute an event or its complement, and justify the choice
Recognise phrasings that signal a complement, and rewrite them
Chain the complement with the two-event rule to answer questions about neither of two events occurring
Identify the one probability that equals its own complement

Where it usually goes wrong

"The complement of the even numbers is the odd numbers." Only if the event was all the even numbers. The chapter's event holds four of the five evens available, so its complement holds 10 as well as the five odds. Name the complement by exclusion, never by a property.
"The rule needs the outcomes to be equally likely." It does not. The proof uses the value of the sure event and additivity, and nothing else. The chapter demonstrates it inside an equally-likely example, which makes the dependence look real when it is not.
"Not A is a different experiment." Same experiment, same outcome list. Only the subset changes.
"At least one means exactly one." They are different events with different probabilities, and Example 7 computes both — 0.98 for one and 0.11 for the other.
"Neither happening is found by multiplying the two failures." That would need a multiplication rule, and this chapter has none. Neither independence nor conditional probability is defined anywhere from p. 289 to p. 313 — checked against every page image. The route the chapter uses is the complement of the union.
"A probability cannot equal its complement." One can, and exactly one does. Example 5's black-card part is it.
"Subtracting from 1 is a shortcut you take when you are stuck." It is a theorem, and in the envelope question it is the difference between one count and three.
"The complement of an event of probability 0.02 is 0.98, so the complement of an event of probability 0.13 is 0.13 less than 1." Both are right, but students frequently subtract the wrong quantity when two events are in play. Always name which event is being complemented before subtracting.

Questions to check understanding

Given an event's probability, state its complement's — Exercise 14.2 Q9 in its barest form
Exercise 14.2 Q3's third and fourth parts ask for a die outcome at or below one and for a die outcome above 6, where one of the two is the impossible event
Exercise 14.2 Q8 asks among other things for no head and for at most two tails on three tossed coins, both of which are faster as complements
Exercise 14.2 Q15's second part and Q21's second part both ask for neither of two events after the union has been found
Exercise 14.2 Q16 gives the complement of both occurring and asks whether the events can occur together
Miscellaneous Exercise Q1, Q4, Q5 and Q6 are all complement questions in disguise, and Q6 is the one where the direct route is materially harder
Miscellaneous Exercise Q3 and Q7 ask for complements alongside other parts, so the student has to notice which parts are worth reversing
Explain, for a given question, why the complement is the cheaper computation

Examples worth working on the board

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

Ten numbered cards (§14.2.4, p. 301). One card is drawn from ten cards bearing 1 to 10, all equally likely at 1/10, and the chapter takes the event {2, 4, 6, 8}. Verified: four outcomes, so 4/10, which is 2/5.
The complement, written out (§14.2.4, pp. 301–302). The chapter lists it as {1, 3, 5, 7, 9, 10}. Verified: six outcomes, so 6/10, which is 3/5, and 3/5 is what remains after 2/5. Also 4 and 6 total the ten cards. This is the bullet: the complement contains 10, which is even. The event was not the even numbers — it was four particular even numbers — so its complement is five odd numbers together with one even one. A student who names the complement by a property rather than by exclusion gets it wrong, and the chapter chose the example that exposes the error without remarking on it.
The proof (§14.2.4, p. 302). The event and its complement share no outcome and together give the whole sample space. Verified: additivity therefore applies to the pair, and the union is the sure event, whose value is 1 by the second axiom, so the two values total 1. The chapter names the two axioms it uses. The non-negativity axiom is not needed here, and neither is the equally-likely hypothesis — the ten-card setting is illustration, not support.
Example 5 on a deck of 52 (pp. 302–303). Verified: an ace is 4 cards out of 52, so 1/13, and not an ace is 12/13 — which the chapter obtains by subtraction rather than by counting the 48. A diamond is 13 out of 52, so 1/4, and not a diamond is 3/4. A black card is 26 out of 52, so 1/2, and not a black card is also 1/2.
The fixed point (pp. 302–303). Verified, and not remarked on by the chapter: a half is the only value a probability can take that leaves its complement equal to it, since a number equals what remains after it only at a half. The black-card part of Example 5 is that case, and it is worth pausing on because students read the equality as an error.
Example 6 on nine discs (pp. 303–304). 4 red, 3 blue, 2 yellow. Blue is 3 out of 9, so 1/3, and not blue is 2/3. Verified by two routes: subtracting 1/3 from 1, and counting the 6 discs that are not blue out of 9. The two routes agreeing is the cleanest possible evidence that the rule is not a trick.
Example 7, chained with the two-event rule (pp. 304–305). With 0.05, 0.10 and 0.02 for one candidate, the other, and both. Verified: neither qualifying is the complement of at least one qualifying, so it is 1 less 0.13, giving 0.87 — and reaching it needs the rule exchanging a complemented union for an intersection of complements. At least one failing is the complement of both qualifying, so it is 1 less 0.02, giving 0.98. Two different complements in one question, taken against two different events.
Three letters, three envelopes (Miscellaneous Exercise Q6, p. 311). Each envelope receives exactly one letter, at random, and the question asks for at least one letter reaching the right envelope. Verified: there are 6 arrangements in all; the arrangements putting nothing in its right place are exactly 2, so the complement is worth 2/6, and the answer is 4/6, which is 2/3. Enumerating all six is quick and makes the complement's advantage obvious — the direct count needs three cases, the complement needs one.
Two sections of a class (Miscellaneous Exercise Q5, p. 311). 100 students split into sections of 40 and 60, with two named students among them. Verified: the pairs from within one section number 780 plus 1770, so 2550, out of 4950 pairs in all, giving 17/33 for the same section; and the different-section case is what remains, namely 16/33. The two parts are a complementary pair and the second should never be computed directly.
More complement triggers (Exercise 14.2 and the Miscellaneous Exercise). Q9 gives an event worth 2/11 and asks for its complement, so 9/11. Q15's second part asks for neither of two events with a union of 5/8, so 3/8. Q16 gives 0.25 for the complement of both occurring and asks whether the two can occur together — verified: both occurring is then worth 0.75, which is not zero, so they can. Q21's second part asks for a student on neither activity, where the either-activity value is 19/30, so 11/30. Miscellaneous Q1's second part asks for at least one green marble among 5 drawn from 10 red, 20 blue and 30 green, which is 1 less the chance that all 5 avoid the 30 green. Miscellaneous Q3's third part asks for not a 3 on a die whose faces carry 1, 1, 2, 2, 2 and 3, so 5/6. Miscellaneous Q4 asks for not winning in a lottery of 10,000 tickets with ten prizes, which for a single ticket is 999/1000.

Figures to have open

Ten numbered cards, four of them liftable, so the complement is visible as what stays behind. An added device; §14.2.4 prints no picture at all, verified against the page images of pp. 301–303.
A box labelled S with one region shaded and the remainder shaded differently. Standard schematic. The chapter's only numbered figure, Fig 14.1 on p. 301, sits on the same page as this section but belongs to the addition-rule topic and shows a different situation.
An enumeration panel for the six letter-to-envelope arrangements, with each letter's correct envelope marked so fixed places are visible. Not in the book.
No photograph or textbook artwork is required.

Where this sits in the book

NCERT Class XI Mathematics, Chapter 14 "Probability", §14.2.4 "Probability of event 'not A'", pp. 301–302, comprising the ten-card illustration and the proof
The scoping sentence at p. 302 announcing that the examples and exercises which follow assume equal likelihood unless stated otherwise
Examples 5 and 6, pp. 302–304; Example 7, pp. 304–305
Exercise 14.2 Q3, Q8, Q9, Q15, Q16 and Q21, pp. 306–308
Miscellaneous Exercise Q1, Q3, Q4, Q5, Q6 and Q7, pp. 310–311
The Summary, p. 312, whose final bullet states the rule
The counting chapter earlier in this book supplies the selection notation the Miscellaneous Exercise questions need — a deliberate cross-reference out of this chapter