Assuming the outcomes are equally likely recovers the counting rule

What to assume they know

• Three conditions on a function, in place of a recipe for counting — the three axioms, and that the one-point values determine every other value
• Events that cannot both happen, and events that between them must — one-point events never overlap, so additivity always applies to them
• The two extreme cases, and the difference between one outcome and many — an event's size is its count of sample points
• Adding fractions over a common denominator, including twenty-eighths
• Selections and arrangements from the counting chapter earlier in this book, in the notation this chapter uses for them
• Factorials, and that four objects can be ordered in 24 ways

What they should be able to do

• Compute an event's probability by adding the values on its outcomes, for an assignment that is not uniform
• Justify that addition by naming the axiom it rests on and the fact that one-point events cannot overlap
• Derive that a uniform assignment on n outcomes must give each outcome 1/n
• State the counting rule together with the hypothesis it requires
• Produce a case where the counting rule returns the wrong number, and explain which hypothesis failed
• Apply the counting rule where sizes must themselves be counted by selection or arrangement
• Choose an outcome list for a loaded object so that the uniformity hypothesis becomes true of it

Where it usually goes wrong

• "Outcomes are equally likely because they are outcomes." It is a hypothesis about the experiment, and the chapter's own two-toss assignment denies it while remaining perfectly legitimate.
• "The counting rule is what probability means." It is a theorem of §14.2.2 resting on a hypothesis stated one line earlier. Where the hypothesis fails the rule fails with it, and the chapter's twenty-eighths example shows it failing by a wide margin.
• "Four outcomes, so each has probability one quarter." Only under uniformity. In the chapter's own weighted example the four values are 7, 4, 8 and 9 twenty-eighths.
• "HT and TH must be equally likely, since both give one head." Not under the chapter's assignment, where they differ by a factor of two. Sameness of description is not sameness of value.
• "A die showing 1, 1, 2, 2, 2, 3 gives each number one chance in three." There are three distinct numbers and six interchangeable faces. Choose the faces as the outcome list and the hypothesis is true of it; choose the numbers and it is false.
• "Favourable means desirable." It means belonging to the event being measured. An event describing a defective pen has favourable outcomes.
• "The denominator counts the events." It counts the outcomes. A four-outcome list carries sixteen events, and none of them is ever the denominator.
• "If you can list the outcomes, you can count them by hand." Example 10's denominator is the number of 7-card hands from 52, which nobody lists. The counting chapter's selection notation is doing the work.
• "Uniformity is an approximation you make to simplify." It is a claim that can be true or false of a physical set-up, and the chapter tells you from p. 302 onward that it is assuming it deliberately rather than deriving it.

Questions to check understanding

• Compute an event's probability from a non-uniform table of outcome values
• Exercise 14.2 Q2, Q3 and Q8 are the plain counting drills — a coin tossed twice for at least one tail, a die thrown for five described events, and three coins tossed for nine described events
• Exercise 14.2 Q4 asks for the size of a 52-card sample space before any probability, then for an ace of spades, an ace, and a black card
• Exercise 14.2 Q5 pairs a fair die with a two-faced coin marked 1 and 6, and asks for two particular totals — a case where the combined list has to be built before it can be counted
• Exercise 14.2 Q7 tosses a fair coin four times with a gain of Re 1 per head and a loss of Rs 1.50 per tail, and asks how many distinct amounts are reachable and with what probability each — the outcome list and the value list are different sizes here, which is the point of the question
• Exercise 14.2 Q10 draws a letter from a 13-letter word and asks for a vowel and for a consonant
• Exercise 14.2 Q11 sets a lottery choosing six different numbers from 1 to 20, with a hint that order does not matter
• Miscellaneous Exercise Q9 forms 4-digit numbers above 5,000 from the digits 0, 1, 3, 5 and 7 and asks for divisibility by 5, once allowing repeats and once not
• Miscellaneous Exercise Q10 asks for a four-wheel lock with digits 0 to 9 and no repeats

Examples worth working on the board

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

• Three pens off a machine (§14.2.1, p. 298). Each pen is recorded as good or bad, so S = {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG} and every outcome is given 1/8. Two events: exactly one bad pen, and at least two bad pens. Verified: exactly one bad means one B and two Gs, which is {BGG, GBG, GGB}, three outcomes, giving 3/8. At least two bad means the three outcomes with two Bs together with the all-bad outcome, which is {BBG, BGB, GBB, BBB}, four outcomes, giving 4/8 and so 1/2. The chapter prints both sets and both answers.
• The same coin, twice, weighted unevenly (§14.2.1, p. 298). The outcome list is the familiar four, but the values assigned are 1/4 to HH, 1/7 to HT, 2/7 to TH and 9/28 to TT. Verified: over twenty-eighths these are 7, 4, 8 and 9, totalling 28, so the assignment passes both tests of the previous topic and is legitimate. Note that the two single-head outcomes carry different values — 1/7 against 2/7 — so they are not interchangeable, and nothing in the axioms forbids it.
• The event both tosses agree (§14.2.1, pp. 298–299). E = {HH, TT}. Verified: 7/28 plus 9/28 is 16/28, which is 4/7. The chapter prints this. Verified, and this is the whole argument of the topic and is not printed: the counting rule would have offered 2 outcomes out of 4, that is 1/2, and 4/7 is not 1/2. The counting rule is simply wrong here, and it is wrong because its hypothesis fails, not because anything was miscalculated.
• The event exactly two heads (§14.2.1, p. 299). F = {HH}, so its value is 1/4 outright. Verified: a one-point event's probability is just the number sitting on its outcome, with no addition needed.
• The forcing argument (§14.2.2, p. 299). Suppose every outcome of an n-member list carries the same value p, with p between 0 and 1 inclusive — read off the page image. The n values total 1, so n copies of p total 1. Verified: p is then 1/n and there is no second possibility; uniformity does not merely suggest the value, it determines it.
• The counting rule (§14.2.2, p. 299). With the list of size n and the event of size m, the probability is m/n, which the chapter also writes out as the favourable outcomes over the possible ones. Verified: this is the previous bullet's 1/n added m times.
• Where the hypothesis is safe (§14.2.4 onward, pp. 302–305). Example 5 draws one card from 52: 13 are diamonds and 26 are black, giving 1/4 and 1/2. Example 6 draws one disc from a bag holding 4 red, 3 blue and 2 yellow, giving 4/9, 3/9 and 2/9. Verified: 4 + 3 + 2 is 9, and the three values total 1, as they must for a family that splits the list. The chapter records from p. 302 onward that its remaining examples assume equal likelihood unless it says otherwise.
• When the sizes need counting (pp. 305–310; note that the Miscellaneous Examples do not start until p. 308, so the first of these is still in the body of §14.2). Example 8 forms a two-person committee, drawing from a group of 2 men together with 2 women; the list has 6 selections. Verified: no man in 1 way, one man in 4, two men in 1, and 1 + 4 + 1 is 6, so the three values are 1/6, 2/3 and 1/6 and they total 1. Example 9 has four cities visited in some order, so 24 orders; the chapter computes A before B as 12 orders and A before B before C as 4. Verified: half the orders put A first of the pair, so 12; and fixing the relative order of three of the four leaves 24 divided by 6, so 4. Verified, and left to the student by the chapter: A first with B last gives 2 orders and so 1/12; A in first or second place gives 12 orders and so 1/2; A immediately before B gives 6 orders and so 1/4. Example 10 draws 7 cards from 52 and gets 1/7735 for four kings and 9/1547 for exactly three. Verified: 1547 times 5 is 7735, so 9/1547 restates as 45/7735; adding the 1/7735 for four kings gives 46/7735, which is the chapter's answer for at least three. The addition is legitimate because three kings and four kings cannot both happen. Example 12 places three of five relay teams, giving 60 orders; ABC in that order is 1 of them, and the same three in any order is 6. Verified: 5 times 4 times 3 is 60, and 6/60 is 1/10.
• A die with repeated faces (Miscellaneous Exercise Q3, p. 311). Two faces carry 1, three carry 2, one carries 3. Verified: the six faces are interchangeable and the three values are not, so the list to use is the faces. Then 2 comes up on three faces, giving 1/2, and 1 or 3 comes up on the other three, also giving 1/2. Counting the three distinct values instead would have given 1/3, which is wrong.

Figures to have open

• Eight cells for the three-pen outcome list, each stamped with its value. Not in the book; §14.2.1 prints the outcomes and their values as two aligned text rows without a rule or caption, so it cannot be cited as a numbered table.
• Four cells for the two-toss list, drawn twice with different stamps. The contrast between the two stampings is the topic's central image.
• A die net showing the repeated faces of the Miscellaneous Exercise question. Standard schematic.
• No textbook figure is needed. §14.2.1 and §14.2.2 print no picture at all, verified against the page images of pp. 298–299; the chapter's only numbered figure, Fig 14.1 on p. 301, belongs to the addition-rule topic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 14 "Probability", §14.2.1 "Probability of an event", pp. 298–299, comprising the three-pen experiment and the weighted two-toss experiment
• §14.2.2 "Probabilities of equally likely outcomes", p. 299, comprising the forcing argument and the counting rule
• The scoping sentence at p. 302 announcing that the remaining examples assume equal likelihood
• Examples 5, 6 and 8, pp. 302–305; Miscellaneous Examples 9, 10 and 12, pp. 308–310
• Exercise 14.2, pp. 305–307; Miscellaneous Exercise Q3, Q9 and Q10, p. 311
• The Summary, p. 312, which states the counting rule using the sizes of the event and of the sample space
• The counting chapter earlier in this book supplies the selection and arrangement notation used from Example 8 onward — a deliberate cross-reference out of this chapter