Three conditions on a function, in place of a recipe for counting

What to assume they know

• Events that cannot both happen, and events that between them must — two events cannot both occur exactly when their intersection is empty
• The two extreme cases, and the difference between one outcome and many — the sure event, the impossible event, and one-point events
• The idea of a function with a stated domain and a stated set of values, from the chapter on relations and functions earlier in this book
• The power set of a finite set, and that it is much larger than the set
• Adding and comparing fractions with unlike denominators, and adding decimals
• Reading a closed interval of real numbers as a set of permitted values

What they should be able to do

• State the three conditions the chapter imposes, and say for each what it rules out
• Identify the domain of the probability function as the collection of events rather than the collection of outcomes
• Derive that the impossible event gets the value 0, using only the third condition
• Test a proposed assignment on a finite outcome list against both requirements, and name which requirement a rejected assignment fails
• Exhibit more than one admissible assignment for the same experiment, and explain why this is not a contradiction
• Derive the upper bound on any event's probability from the three conditions, rather than assuming it
• Explain why the values on the one-point events determine every other value

Where it usually goes wrong

• "Probability means favourable cases over total cases." Not in this section. That rule needs an extra assumption the axioms do not make, and the next topic is where it is recovered — as a consequence, under a hypothesis, rather than as a definition.
• "The axioms tell you the chance of a head." They do not, and the chapter says so by exhibiting an unlimited family of admissible answers for one coin. Anything further has to come from the experiment, not the mathematics.
• "A valid assignment must be positive everywhere." Assignment (b) gives five outcomes the value 0 and is accepted. The chapter's own commentary on row (a) describes the entries as positive and below 1 while its commentary on row (b) accepts a 1 and five 0s — the two readings of the same condition do not agree, and the stated condition, which allows both ends, is the one to trust.
• "Probability 0 means the outcome is not there." Under assignment (b) five outcomes carry 0 and remain in the outcome list. The chapter proves the empty event has value 0; it never proves the converse.
• "An entry above 1 is fine as long as the others are small." The one-point values total 1 and none is negative, so none can exceed 1. Row (d) is rejected on that entry alone.
• "If every entry is a sensible probability, the assignment is valid." Row (e) is exactly that and still fails. Both tests have to be run.
• "The third condition lets you add any two probabilities." Only two that cannot both occur. Everything the chapter later has to prove about adding probabilities exists because that restriction is real.
• "P(E) ≤ 1 is one of the three axioms." It is not among them. The chapter builds it into the declaration of the function's values, and it can be got out of the three conditions instead.
• "Negative probability is just a very small chance." The first condition forbids it outright, and two entries of row (c) are rejected on that ground.

Questions to check understanding

• Decide whether a proposed assignment is admissible, and name the requirement it fails — Exercise 14.2 Q1 sets five rows on a seven-outcome list, one of them a uniform seventh, one summing to 2.8, one carrying two negative entries, and one ending in a value above 1
• Derive the value of the impossible event from the additivity condition
• Given a partial assignment, find the value that completes it
• State the three conditions and give an assignment failing exactly one of them
• Exhibit two different admissible assignments on the same outcome list
• Explain why the additivity condition cannot be applied to a stated overlapping pair

Examples worth working on the board

Values marked verified are worked out here against the printed page; the chapter states one reason for each rejection and computes almost nothing else.

• The declaration (§14.2, p. 296). Probability is set up as a function taking every subset of the sample space as an input, and returning a number in the closed stretch from 0 to 1. Both the lower and the upper end are included; this was read off the page image, not the extracted text.
• The three conditions (§14.2, p. 296). Non-negativity for every event, with the comparison inclusive — zero is permitted; the value 1 for the sure event; and additivity across a pair that cannot both occur. All three were read off the page image.
• The empty event's value (§14.2, p. 296). The chapter's argument: take the second event of the additivity condition to be the empty one. It shares nothing with the first, and unioning it changes nothing, so the first event's probability equals itself plus the empty event's. Verified: subtracting gives zero, and the step uses only the third condition — neither of the other two is needed.
• What the conditions force on a finite list (§14.2, p. 296). For an outcome list of n members: each one-point value lies between 0 and 1 inclusive, the n one-point values total 1, and any event's value is the sum over the outcomes it holds. Verified, and not printed: this makes the n one-point values the only free data — every one of the 2ⁿ events then has a determined value, so an eight-outcome experiment needs 8 numbers to pin down 256. The chapter never counts its events, so the figure 256 must be presented as added here.
• One coin, more than one answer (§14.2, pp. 296–297). The chapter first assigns each face 1/2 and checks both requirements. It then assigns 1/4 and 3/4 and checks them again. Verified: each pair sums to 1 and each entry sits in range, so both pass. The chapter then generalises to p against 1 − p for any p from 0 to 1 inclusive, giving infinitely many admissible assignments for one coin.
• Example 4 — five candidate assignments (§14.2, p. 297). Six outcomes.
  • (a) 1/6, 1/6, 1/6, 1/6, 1/6, 1/6
  • (b) 1, 0, 0, 0, 0, 0
  • (c) 1/8, 2/3, 1/3, 1/3, −1/4, −1/3
  • (d) 1/12, 1/12, 1/6, 1/6, 1/6, 3/2
  • (e) 0.1, 0.2, 0.3, 0.4, 0.5, 0.6

Verified: (a) totals 1 and every entry is in range, so it passes. (b) totals 1 and every entry is in range, so it passes as well. (c) carries two negative entries. (d) carries an entry above 1. (e) totals 2.1.

• Reading the whole row rather than the first fault (§14.2, pp. 297–298). Verified, and none of it printed: (c) also totals 7/8, so it fails both tests, not just the range test; and (d) also totals 13/6, so it fails both too. Only (e) is a clean single failure — every entry of (e) is a legitimate probability on its own, and the row is rejected purely because the six of them total more than 1. That contrast is the sharpest way to show the two tests are independent, and the chapter gives exactly one reason per row and does not make it.
• The upper bound, derived (§14.2, p. 296). Verified, and not printed as a derivation: the sure event splits into any event and its complement, which cannot both occur, so their values total 1; since neither is negative, neither can exceed 1. The chapter asserts the upper end by declaring the function's values and never derives it from the three conditions.

Figures to have open

• A schematic of the probability function: a column of events on the left, a number line from 0 to 1 on the right, arrows landing inside it. Not in the book; §14.2 prints no picture, verified against the page images of pp. 295–298.
• A box labelled S with two non-overlapping shaded regions, redrawn with the regions overlapping, to make the third condition's restriction visible. Standard schematic.
• A five-row table of the candidate assignments with two verdict columns, one per test. An added arrangement of the chapter's printed table.
• A dated strip for the Historical Note. The names and years come from p. 313 and need no artwork from the book.
• Fig 14.1 on p. 301 is the chapter's only numbered figure and is not needed here.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 14 "Probability", §14.2 "Axiomatic Approach to Probability", pp. 295–298, comprising the declaration of the function, the three axioms, the derivation for the empty event, the three consequences on a finite outcome list, the Note on one-point events, the coin assignments and Example 4
• Exercise 14.2 Q1, pp. 305–306
• The Historical Note, p. 313, which credits the axiomatic theory to Kolmogorov and dates his book to 1933
• The Summary, p. 312, whose probability bullet restates the three consequences rather than the three axioms