The rules conditional probability inherits, including addition on a restricted space

What to assume they know

• The definition of a conditional probability from the previous topic, including its side condition
• The addition rule for two events on an unrestricted space, from Class XI
• An event and its complement, and the fact that they exhaust the space between them
• Union and intersection distributing across each other, and De Morgan in outline
• Disjoint events, and that the probability of their union is a plain sum
• Reading a probability statement in which two events are combined on one side of the bar

What they should be able to do

• State the claim that a conditional probability is itself a probability, and say what would have to fail for that to be false
• Prove that the whole space, given an event, comes out certain, and that the event given itself does too
• Write the addition rule with a fixed ground on the right of the bar, and identify the term that gets subtracted
• Reproduce the proof of that rule, naming the set identity the second line uses
• Specialise the rule to two events with no shared outcome and say why the third term disappears
• Derive the complement rule from the first property plus the disjointness of an event and its opposite
• Recognise which statements about a restricted ground are licensed and which are not
• Evaluate a compound conditional in the shape of Exercise 13.1 Q11, where a union and an intersection are both conditioned on a third event
• Explain why the later theorems of the chapter never re-prove any of this

Where it usually goes wrong

• "These are three more formulas." They are one claim in three parts: conditioning yields a probability. Present them as a certificate, not a list, and the section takes four minutes instead of ten.
• "The addition rule needs the two events to be disjoint." It does not. The general form subtracts the overlap; the disjoint form is the special case where there is nothing to subtract. Students who learn only the short form silently double-count.
• "You can put the union on the right of the bar too." Nothing on Part II pp. 408–409 says anything about that. All three results hold one fixed ground at a time.
• "The event given itself being one is a triviality not worth a line." It is, and that is exactly why the chapter proves it — because the restricted assignment has to be shown to total one on its own ground before anything can be built on top of it.
• "The complement rule is a new fact." It is Property 1 and the disjoint form of Property 2 put together. Deriving it in three lines is cheaper than memorising it and shows the two earlier results doing work.
• "A conditional probability could come out above one." It cannot, but this section never says so — the bound appears only in the Summary. If a student asks, the reason is that the numerator counts a subset of what the denominator counts. Flag it as a fact the body of the chapter leaves out.
• "The distributive step is the one about union spreading over intersection." It is the other one. The proof intersects a union with a fixed set. The chapter's own bracketed label is the wrong way round, so a student who trusts it will look for the wrong identity.

Questions to check understanding

• State the addition rule with a fixed conditioning event and identify each of its four terms
• Prove that an event given itself has probability one, from the definition
• Reproduce the proof of the addition rule and name the set identity it uses
• Derive the complement rule from the two earlier results
• Given three events on a small space, compute a union conditional both by direct counting and by the rule, and check they agree — the form of Exercise 13.1 Q11
• Say which of a list of proposed identities are licensed by this section, with at least one distractor that moves the conditioning event
• Explain why the shorter form of the addition rule is not the general one

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.

• Property 1 (§13.2.1, Part II p. 408). Two statements in one line: the whole space given an event, and that event given itself, both come out at one. Each proof is a single substitution — the intersection of the whole space with the event is the event, and the intersection of the event with itself is the event — so both fractions collapse to the same thing over itself. Run both simultaneously; they are the same two lines.
• Why it is worth proving. The chapter does not say. This is an added framing: the result is what certifies that the restricted assignment is a probability at all. Without it there is nothing to stop conditioning producing numbers that fail to total one, and every later argument in the chapter — the totalling over a partition, the reversal of a conditional — would need its own ground-up justification.
• Property 2 (Part II pp. 408–409). Two events combined by union on the left, with a third event fixed on the right, equals the two conditionals added and the conditional of their intersection taken off once. Note the shape: it is the ordinary addition rule with every term carrying the same fixed right-hand side.
• The proof, four lines (Part II p. 409). Put the union over the fixed event into the definition; rewrite the numerator by distributing the fixed event across the union; apply the ordinary addition rule to the two pieces, whose own overlap is the triple intersection; then split the single fraction into three and read each back as a conditional. The only step that is not bookkeeping is the second.
• The set identity in that second step (Part II p. 409). What is used is that intersecting a union with a fixed set gives the union of the two intersections. Show it as a shaded diagram before the algebra; it is the one place a student can get lost. The chapter's own bracketed justification names the two operations in the opposite order from the identity it is justifying — see Notes, and do not read the printed phrase aloud.
• The disjoint case (Part II p. 409). If the two events share no outcome then the conditional of their intersection is nought and the rule shortens to a plain sum. The chapter prints this twice, once as a display line before the proof and once as a consequence after it. Verified: the vanishing term is the conditional of the empty event, whose numerator is the probability of the empty set.
• Property 3 (Part II p. 409). An event's opposite, on a fixed ground, is one minus the event on that same ground. The proof is three lines: the whole space on that ground is one by Property 1; the whole space splits into the event and its opposite; those two are disjoint, so Property 2 in its short form applies. This is the cleanest illustration in the chapter of one printed result being used to get the next.
• What is not licensed. An added section, and it is the one a student needs. All three results hold with a single fixed event on the right. Nothing here says anything about combining two conditionals with different grounds, and nothing here lets a union or an intersection be formed across the bar. Show a crossed-out non-rule beside the three real ones.
• Exercise 13.1 Q11 (Part II p. 414). A fair die with three named events: the odd faces, a two-element set, and a four-element set. Part (iii) asks for the union of the first two and then their intersection, both conditioned on the third. Verified from the chapter's own sets: the union of the first two events is four faces, three of which lie in the conditioning event of four, giving three quarters; their intersection is the single face three, which lies in the conditioning event, giving one quarter. Verified by the other route, which is the point of the item: the two separate conditionals are two quarters and two quarters, their intersection conditional is one quarter, and two quarters plus two quarters less one quarter is three quarters. The two routes agree, which is Property 2 checked on a case a student can count by hand.
• Exercise 13.1 Q11, parts (i) and (ii) (Part II p. 414). Verified: the odd faces given the two-element set is one half, and the reverse is one third; the odd faces given the four-element set is one half, and the reverse is two thirds. Use one of these pairs again for the asymmetry point if the previous topic's example has faded.
• The Summary's version (Part II p. 437). Bullet two of the Summary prints the addition rule with a fixed ground, matching Property 2, and the complement rule, matching Property 3 — and prints a two-sided bound on a conditional probability that §13.2.1 never states, while omitting Property 1 altogether. Both facts were read off the printed page. A student revising from the Summary meets a bound with no proof behind it and loses the result the other two are built on.

Figures to have open

• Two shaded region diagrams for section 5: a union intersected with a fixed set, and the union of the two separate intersections, drawn at the same size and in the same position so the student sees one shape twice rather than two shapes. The chapter draws no figure at all for this proof; Fig 13.3 on Part II p. 420 is the nearest thing it prints and it belongs to a different argument.
• A colour convention held across sections 4 to 9: the conditioning event always the same colour, the events being measured always another. If the explanation changes the colour when the event changes role, the fixed-ground point is lost.
• A single die-face row for section 10, with the three exercise events drawn as three bands beneath it, so the union and the intersection can both be read off one picture.
• A two-column comparison for section 11 built with the repo's DataTable component, listing what the section proves against what the Summary prints. Peers here must share one fitSize; the Summary column has longer lines and will otherwise typeset smaller and read as subordinate.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 13 "Probability", §13.2.1 Properties of conditional probability, Part II pp. 408–409
• Property 1 with both proofs, Part II p. 408
• Property 2, its statement and its disjoint special case, Part II p. 408; the four-line proof, Part II p. 409
• Property 3 with its proof, Part II p. 409
• §13.1's reference to the addition rule as earlier learning, Part II p. 406
• Exercise 13.1, question 11, Part II p. 414
• Summary, second bullet, Part II p. 437