Multiplying along a chain of dependent draws, and extending it past two events

What to assume they know

• The definition of a conditional probability and its side condition, from the first module
• Intersection of events, and the fact that it does not care about order
• The three properties of a conditional probability from §13.2.1
• Counting a deck of fifty-two cards by suit and rank, from Class XI
• Multiplying and cancelling fractions with large denominators
• The idea of drawing from a container with and without putting the item back

What they should be able to do

• Derive the product form from the definition of a conditional probability, in one line, in both orders
• Say why the two orders give the same value and what each needs to be nonzero
• Read the shorthand in which an intersection of two events is written by juxtaposition
• Identify, for a two-stage draw, which factor is unconditional and which is not
• Recompute a second-stage probability after the first stage has changed the contents of the container
• Extend the product to three events and write the condition on the third factor in both of the chapter's notations
• Apply the same pattern to four or more stages without re-deriving it
• Work a three-card draw and reduce the resulting fraction correctly
• Recognise the same rule appearing inside the later theorems of the chapter, where it is used but not renamed

Where it usually goes wrong

• "The second factor is just the probability of the second event." Only when the first draw changed nothing. In Example 8 the container has fourteen balls at the second stage, and using fifteen is the single most common wrong answer in the whole section.
• "You have to count the intersection." You never do. That is the point of the rule: the intersection is expensive to count directly and cheap to build as a product.
• "The two orders give different answers." They give the same value, because the intersection is symmetric. They are different computations — one needs the first probability and one conditional, the other needs the second probability and the other conditional — and in any given question one of the two is available and the other is not.
• "Without replacement means the second factor is a guess." It is exactly determined. Say out loud what the container holds after the first draw, then read the second factor off that sentence, as the chapter does.
• "For three events the last condition is only the event just before it." It is both earlier events at once. A student who conditions only on the immediately preceding stage will get Example 9 wrong at the third factor.
• "Side-by-side event names mean a product of probabilities." They mean the intersection. The shorthand is introduced on Part II p. 415 and used inside a conditional bar on Part II p. 416, where mistaking it for a product produces nonsense.
• "Each label on a tree branch is a probability of an outcome." Depends whose tree. In this chapter's own drawing the second-level labels are outcome probabilities; in the chain figure this topic needs, the edge labels are conditionals. Say which convention is shown every time a tree appears.
• "§13.3 must have its own exercise somewhere." It does not. Its practice items sit inside Exercise 13.2, which is printed after the next section. A student working section by section will meet the multiplication questions under an independence heading and assume every one of them is about independence.

Questions to check understanding

• Derive the product form from the definition of a conditional probability, in both orders
• Given a container and two successive draws without replacement, state what the container holds at the second stage before computing anything
• Compute the chance of a two-stage outcome, once with replacement and once without, and explain which factor differs
• Write the three-event product and state the condition on the third factor
• Work a three-card draw from a full deck and reduce the answer fully — the form of Example 9 and of Exercise 13.2 Q3
• Read a chain figure and say for each label whether it is a conditional or a running product
• Answer the king-and-queen question the section opens with, stating which of the two readings of the wording you have taken

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 of §13.3 (Part II p. 415). Three things happen in a paragraph: the intersection is read aloud as both events happening together; a shorthand is introduced in which the two event names are written side by side to mean it; and a motivating question is posed about drawing a king and a queen from a deck one after the other. The chapter poses that king-and-queen question and never answers it. Do not leave it hanging in the explanation either — either answer it or cut it. Verified, if you answer it: drawing a king then a queen without putting the first card back has probability four over fifty-two times four over fifty-one, which is four over six hundred and sixty-three; allowing either order doubles it to eight over six hundred and sixty-three. The chapter supplies neither number and does not say which of the two readings it means, so if this is shown it must be labelled as working added here.
• The derivation (Part II p. 415). Take the definition, multiply both sides by the probability of the conditioning event, and read the result. Then do it again with the roles swapped, using the fact that an intersection does not care about order. Two lines, two labels, and the combined statement carries both nonzero conditions. Show the multiplication happening; it is the entire section.
• Example 8 (Part II pp. 415–416), the urn. Ten black and five white balls, two drawn in succession with nothing put back, both wanted black. First factor ten over fifteen. Then the chapter says in prose what the container now holds — nine black among fourteen — and reads the second factor straight off it. Verified: ten over fifteen times nine over fourteen is three sevenths. Note: the chapter leaves the first factor unreduced as ten over fifteen rather than two thirds, which is the right pedagogical choice here because fifteen is the number the second factor has to be built from. Keep it unreduced for the same reason.
• The move from two events to three (Part II p. 416). Printed as a run-in heading in bold italic rather than as a numbered subsection, which is why a reader scanning for section numbers will miss it. The third factor's condition is that both earlier events happened, and the chapter writes that condition twice in one display line, once with brackets and once in the side-by-side shorthand. Show both at that moment; it is the only place the shorthand is exercised.
• Example 9 (Part II pp. 416–417), three cards drawn in succession from a full deck with nothing put back: two kings, then an ace. The chapter narrates each denominator shrinking — fifty-two, then fifty-one, then fifty — and each numerator separately: four kings, then three, then four aces. Verified: the product is forty-eight over one hundred and thirty-two thousand six hundred, which reduces to two over five thousand five hundred and twenty-five. That reduction divides both parts by twenty-four; do it, because a student who cannot see where the printed answer came from will assume it was looked up.
• The extension sentence (Part II p. 416). One line saying the same pattern runs to four events or more. No proof, no example. That is honest and it is enough — the induction is obvious once the three-event case is shown.
• Exercise 13.2 Q2 (Part II p. 421). Two cards drawn from a full deck with nothing put back, both wanted black. Verified: twenty-six over fifty-two times twenty-five over fifty-one, which is twenty-five over one hundred and two. This is Example 8 with a deck instead of an urn, and it is the cleanest single item to set as practice.
• Exercise 13.2 Q3 (Part II p. 421). Fifteen oranges, twelve of them sound, three drawn without replacement and all three wanted sound. Verified: twelve over fifteen times eleven over fourteen times ten over thirteen, which is forty-four over ninety-one. A three-factor chain in the wild.
• Exercise 13.2 Q13 (Part II p. 422). The contrast case: ten black and eight red balls drawn with replacement. Verified: both red is eight over eighteen squared, sixteen over eighty-one; black then red is ten over eighteen times eight over eighteen, twenty over eighty-one; one of each in either order is twice that, forty over eighty-one. Run this immediately after Example 8. The same rule applies, and the conditional factor simply happens to equal the unconditional one — which is the bridge into the next topic.
• The tree as the picture (Fig 13.1 and Fig 13.2, Part II p. 412). The chapter draws its only tree back in §13.2 and never redraws one here, even though this is the section a tree belongs to. Reuse the shape, not the content. Warning read off the printed page: in Fig 13.2 the numbers on the second level are the leaf probabilities, not the branch chances. A chain figure drawn for this topic must put the conditional chance on the edge and the running product at the node, and must say that it differs from the chapter's own drawing.

Figures to have open

• One urn, drawn twice at the same size and position, holding fifteen balls and then fourteen. The removed ball should leave the frame rather than vanish, so the fourteen is seen to be caused. The chapter draws no urn; this is added here.
• A deck strip for section 9 showing three cards leaving in order, with the remaining count updating beneath. Sized once for all three panels with a shared fitSize, so the third panel does not typeset smaller than the first.
• A chain diagram for section 11 built with the repo's Network component: edges carry conditional chances, nodes carry the running product. It must not reproduce Fig 13.2's labelling convention, and the caption should say so.
• A single reduction movement for the three-card product, showing both parts divided by twenty-four. Without it the printed answer looks conjured.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 13 "Probability", §13.3 Multiplication Theorem on Probability, Part II pp. 415–417
• The shorthand for an intersection and the opening question, Part II p. 415
• The derivation in both orders and the combined statement, Part II p. 415
• Example 8, Part II pp. 415–416
• The run-in heading extending the rule past two events, and the extension sentence, Part II p. 416
• Example 9, Part II pp. 416–417
• Exercise 13.2 questions 2, 3 and 13, Part II pp. 421–422
• Fig 13.1 and Fig 13.2, Part II p. 412
• Summary, third bullet, Part II p. 437