Being told one event happened shrinks the sample space you count against

What to assume they know

• Sample space, outcome and event from Class XI, and an event written as a set
• Equally likely outcomes, and probability as a count of favourable cases over a count of all cases
• The axiomatic treatment of probability, which the chapter's opening page names as prior learning
• Union, intersection and complement of sets, and the empty set
• Reading a set of ordered pairs or ordered triples as a list of outcomes
• Adding fractions with unlike denominators, and dividing one fraction by another

What they should be able to do

• Describe in one sentence what being told that an event has happened does to the list of outcomes still in play
• Reproduce the chapter's three-coin argument, naming the two events and the four outcomes each contains
• Explain why the outcomes that stay favourable are precisely those in the intersection of the two events
• Write the conditional probability as a ratio of two counts, and then as a ratio of two probabilities, and show that the second follows from the first
• State Definition 1 with its side condition, and say what goes wrong without it
• Read the vertical bar correctly: name which event supplies the new ground and which one is being measured
• Compute both conditionals of a given pair and show that they are in general different numbers
• Apply the definition where the outcomes are not equally likely, using probabilities of outcomes rather than a head count
• Read a tree diagram of a two-stage experiment and recover the outcome list from it

Where it usually goes wrong

• "Conditioning changes the coins." It changes nothing physical. The coins have already landed. What the information changes is which outcomes you are entitled to count, which is a statement about your list, not about the world.
• "The probability of the first given the second is the same as the second given the first." They share a numerator and have different denominators, so they agree only when the two events have equal probability. Exercise 13.1 Q1 makes them two thirds and one third on the same three inputs; show both together.
• "Divide by the probability of the event you want." Divide by the event you were told about. Deriving the formula from the head count fixes this permanently, because in the count form the denominator is visibly the list you are standing in.
• "The condition on the denominator is a technicality." It is the difference between an answer and no answer. Exercise 13.1 Q16 exists to make a student write not defined rather than nought.
• "Conditional probability is a smaller probability." It can be larger, smaller or equal. In Example 5 it equals the unconditional chance; in Example 2 it is smaller; conditioning on a rarer event can push it up. Show one of each.
• "Both children being boys, given one is a boy, is one half." The famous one. There are three surviving pairs, not two, because a girl-then-boy family and a boy-then-girl family are different outcomes and both qualify. The chapter lists all four pairs on Part II p. 410 for exactly this reason.
• "Once outcomes stop being equally likely, the whole method fails." Only the head-counting fails. Example 7 runs the same definition on a space where the outcomes carry two different probabilities.
• "Every branch label on a tree is a branch chance." Not in this chapter's own drawing. Fig 13.2 labels its second level with leaf probabilities. Say so when the figure is shown, or a student will multiply along the path and get a wrong number that looks right.

Questions to check understanding

• Given three probabilities, produce both conditionals of the pair and say why they differ
• Given a small outcome list and two events, write down the reduced list and count inside it
• Recover the probability of both events from a conditional and a single probability, in the shape of Exercise 13.1 Q3
• State the side condition on the definition and say what the answer is when it fails, in the shape of Exercise 13.1 Q16
• Work a conditional on a space whose outcomes are not equally likely, in the shape of Example 7
• Read a two-stage tree and list the outcomes with their probabilities
• Explain, without notation, what information does and does not change about an experiment that has already been run

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 three-coin experiment (§13.2, Part II pp. 406–407). Eight outcomes, each given probability one eighth because the coins are fair. The first event is that at least two heads show, and it holds four outcomes; the second is that the first coin shows tails, and it also holds four. Verified: each therefore has probability one half, the two share exactly one outcome — tails then heads then heads — so the probability of both is one eighth. Conditioned on the second event, only one of its four surviving outcomes is favourable, giving one quarter. Every number here is derived from the chapter's own outcome lists.
• The move that does all the work (Part II p. 407). The chapter says in prose that once you know the second event happened, outcomes that contradict it drop out of consideration, and that you may treat what is left as a fresh experiment in its own right. That sentence is the entire idea; the algebra on the next page is bookkeeping.
• The two forms of the formula (Part II p. 408). First as a ratio of head counts — outcomes in the intersection over outcomes in the given event — and then, after dividing numerator and denominator by the size of the whole space, as a ratio of two probabilities. Show the division; it is three seconds of movement and it removes the commonest memorisation error.
• Definition 1 (Part II p. 408), which is the first of only three numbered definitions in the chapter. It carries the condition that the given event must have nonzero probability. The chapter marks the reason with a bracketed challenge to the reader rather than answering it. Verified: the denominator would be zero, and separately the given event would be empty, so there would be no reduced list to count against at all.
• Example 1 (Part II p. 409). Pure substitution: three probabilities in thirteenths are given and one conditional is wanted. Verified: four over thirteen divided by nine over thirteen is four ninths. Use it as the arithmetic drill, not as the idea.
• Example 2 (Part II pp. 409–410), two children. The outcome list is four ordered pairs of boy and girl. Verified: three pairs contain at least one boy, one pair is two boys, so the answer is one third — not one half, which is what most students will say out loud before the reveal.
• Example 3 (Part II p. 410), ten numbered cards. Verified: seven numbers exceed three, four of those seven are even, so four sevenths. This is the cleanest instance of counting inside the reduced list.
• Example 4 (Part II pp. 410–411), a school of one thousand with four hundred and thirty girls, a tenth of them in Class XII. Verified: the chance of being a girl is nought point four three, the chance of being a Class XII girl is nought point nought four three, and the quotient is nought point one — which is the tenth you were handed, arriving back unchanged. Say that out loud: the worked answer is the input, which is what makes it a good check and a poor puzzle.
• Example 5 (Part II p. 411), a die thrown three times, with two hundred and sixteen outcomes. Verified: the second event pins the first two throws and leaves six outcomes; one of them also has four on the third throw; so one sixth, which is just the chance of a four on a single throw. Worth pausing on — it is the first hint of independence, which module two makes explicit.
• Example 6 (Part II pp. 411–412), a die twice with the two faces summing to six. Verified: five outcomes sum to six, two of them contain a four, so two fifths. Note that the chapter also prints the unconditional chance of a four appearing at least once, eleven over thirty-six, which it then never uses. That spare number is useful precisely as the thing the condition replaces.
• The generalisation paragraph (Part II p. 412). The chapter states that the same definition applies when the outcomes are not equally likely, provided the two probabilities in the fraction are worked out properly. This is where the ratio-of-probabilities form earns its keep, because the ratio-of-counts form becomes wrong.
• Example 7 and the tree (Part II pp. 412–413). Toss a coin; on heads toss again, on tails throw a die. Eight outcomes with unequal probabilities: two of one quarter and six of one twelfth. Verified: the event that at least one tail appears has probability three quarters, the die showing more than four carries probability one sixth, and the conditional is two ninths. Check the arithmetic — a quarter plus six twelfths is three quarters, and a sixth divided by three quarters is two ninths.
• Fig 13.1 and Fig 13.2 (Part II p. 412), the two tree panels. Read off the printed page: the first panel is the bare tree, two branches at the first level and eight leaves at the second, carrying no numbers at all; the second panel is the same tree with numbers added. The numbers on the second level of Fig 13.2 are the leaf probabilities, not the branch chances — one quarter on each of the two head branches and one twelfth on each of the six tail branches, where the chances along those branches are one half and one sixth. This is a real trap and the chapter does not warn about it: multiplying the two labels along a path does not give the leaf. See Notes.
• Exercise 13.1 (Part II pp. 413–415), seventeen items. Worth lifting: Q1, three probabilities given and both conditionals wanted, which makes the asymmetry unavoidable — verified two thirds one way and one third the other; Q3, where a conditional is given and the intersection has to be recovered from it — verified nought point three two, then nought point six four, then nought point nine eight; Q5, where the intersection comes out of the addition rule first — verified four elevenths, four fifths, two thirds; and Q16, whose answer is that the quantity is simply not defined. Q17 is discussed in Notes, because as printed it has no correct option.

Figures to have open

• A single strip of eight outcome cells for the three-coin experiment, reused in sections 2, 3 and 4 so the student watches one object change state rather than three drawings appear. The chapter prints the outcome list but draws no such strip; the layout is added here.
• A two-column build for section 4 that lights the shared outcome in both columns at once. Do not use useGroupAttention to dim one column here — both columns are the subject of the same sentence.
• A fraction-to-fraction transformation for section 6 in which the two counts visibly acquire and then cancel a denominator of eight. This is the one movement the topic cannot do without.
• Five small reduced-list panels for section 10, all sized with a single shared fitSize so that Example 4's long school sentence does not shrink the other four.
• The two tree panels of Part II p. 412. Redraw rather than reproduce, and label the second-level numbers as leaf probabilities. Build with the repo's Network component; a chain of stages is what it is for.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 13 "Probability", §13.1 Introduction, Part II p. 406
• §13.2 Conditional Probability: the three-coin argument, Part II pp. 406–407; the two forms of the formula and Definition 1, Part II p. 408
• Examples 1 to 6, Part II pp. 409–412
• The paragraph extending the definition to outcomes that are not equally likely, and Example 7 with Fig 13.1 and Fig 13.2, Part II pp. 412–413
• Exercise 13.1, Part II pp. 413–415
• Summary, first two bullets, Part II p. 437