Complements: knowing one probability hands you the other

What to assume they know

• Favourable over total: the definition this chapter runs on — the definition as a ratio of counts
• Single-outcome events, and why all of them together come to 1 — that a group of events covering everything without overlap totals 1
• Subtracting a fraction from 1, and subtracting a decimal from 1
• Reading a bar drawn over a symbol as a piece of notation rather than as an operation

What they should be able to do

• Recognise, in a pair of worked answers, that the second event is the denial of the first
• Write the denial of an event using the chapter's bar notation
• Explain why the two probabilities total 1 by pointing at the outcomes rather than at the arithmetic
• Rearrange the total into a subtraction and use it to find one probability from the other
• Compute a probability twice — once by direct count, once by subtraction — and check that the two agree
• Identify the situations where the complement is the cheaper thing to count, particularly events phrased with "at least"
• State the condition under which two events may be treated as each other's denial, and give a situation where it would fail
• Read a probability given in a question as the complement of the one asked for

Where it usually goes wrong

• "The denial of 'at least one head' is 'at least one tail'." Both of those have probability 3/4 and both can happen on the same toss. The denial of "at least one head" is "no head anywhere". This single confusion accounts for most wrong answers on Example 9 and question 24.
• "Any two events that sound opposite are complements." They must cover every outcome and share none. The chapter's tennis match qualifies only because the match must produce a winner; introduce a possible draw and the subtraction is simply wrong.
• "The rule saves work every time." It saves work when the denial is easier to count. In Example 4 the direct count is just as quick, and the chapter does it both ways precisely to show the two agree.
• "A bar over the answer means I should subtract twice." Once is always enough. Questions 5 and 7 look like mirror images — one hands over the event and wants its complement, the other hands over the complement and wants the event — but both are answered by taking the given number from 1, and a student who subtracts a second time to "undo" the reversal turns 0.008 back into 0.992.
• "A bar over a letter is an operation like a minus sign." It is a name. The arithmetic happens afterwards.
• "Complement means the leftover unlikely bit." In Example 4 the complement is the larger part by a long way, at 12/13.

Questions to check understanding

• Given one probability, state the other, as in Exercise 14.1 question 5
• Given the complement, recover the event, as in question 7 — the reversed form that catches reflex subtraction
• Compute a probability both directly and by subtraction and show the agreement
• Any question phrased with "at least", which is the standard board signal that the complement is the intended route
• Explain why two named events may be treated as complementary, which is the reasoning version of the same skill
• Two-part questions where one part is the denial of the other, so that the total can be checked

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data.

• The two patterns (p. 206). From Example 1, head and tail at 1/2 each. Verified: they total 1. From Example 3, a die result above 4 at 1/3 and a result of 4 or below at 2/3. Verified: they total 1 as well. The page then makes the observation that in each pair the second event is the first one denied, and only after that introduces the notation and the rule.
• The notation and the rule (p. 206). A bar over the letter names the denial; the two probabilities total 1; rearranged, one is 1 minus the other. Keep the order the page uses — pattern, then observation, then notation, then rule. An explanation that opens with the formula has thrown away the derivation.
• Example 4 (pp. 207–208). One card from a shuffled pack of 52. Four aces, so verified: P(ace) = 4/52 = 1/13. For a card that is not an ace, the direct count is 52 − 4 = 48, so verified: 48/52 = 12/13. The following remark redoes it as 1 − 1/13 and gets 12/13. This is the chapter's own double-check and it is the best demonstration in the topic, because both routes are short enough to show in full.
• Example 5 (p. 208). A tennis match, its two players being Sangeeta and Reshma; Sangeeta's chance of winning is given as 0.62. Verified: 1 − 0.62 = 0.38 for Reshma. The page treats the two as each other's denial. The assumption is unstated: it works because a tennis match yields a winner, so there is no third outcome. Show this — it is the one place in the chapter where the complement rule is applied to something that is not a count, and the justification is entirely in the background.
• Example 6 (p. 208). Savita and Hamida, leap years set aside, so 365 days treated as equally likely for the second birthday. Verified: different birthdays are favoured by 364 of the 365, giving 364/365; the same birthday is then 1 − 364/365 = 1/365. Note which side the chapter counts directly: the large one, because the small one is awkward to argue and trivial to subtract.
• Example 7's closing note (p. 209). Forty students, 25 girls and 15 boys, one name card drawn. Verified: P(girl) = 25/40 = 5/8 and P(boy) = 15/40 = 3/8, and 1 − 5/8 = 3/8 recovers the second from the first.
• Example 9's closing note (p. 210). Two coins that can be told apart, tossed together. Verified: the four outcomes are HH, HT, TH, TT; at least one head is favoured by three of them, so 3/4; and the denial — no head at all — is favoured by exactly one, so 1/4, and 1 − 1/4 = 3/4. This is the section-10 example and the reason the topic matters: three cases counted, or one case subtracted.
• Exercise 14.1 question 5 (p. 214). Given 0.05. Verified: 0.95.
• Exercise 14.1 question 7 (p. 214). Among three students, the chance that two of them do not share a birthday is given as 0.992. Verified: 0.008 for the shared birthday. Note that the arithmetic here is identical to question 5's — one minus what you were given, both times. What differs is only which member of the complementary pair the question chooses to hand over, and that is a labelling point rather than a computational one. Neither item punishes the reflex.
• Exercise 14.1 question 8 (p. 214). A bag of 3 red and 5 black balls. Verified: red 3/8; not red 5/8, obtainable either by direct count or by subtraction.
• Exercise 14.1 question 21 (p. 216). A lot of 144 ball pens with 20 defective. Verified: Nuri buys with probability 124/144 = 31/36 and declines with 20/144 = 5/36, and the two total 1.
• Exercise 14.1 question 24 (p. 216). A die thrown twice, with a printed hint that this is the same experiment as two dice thrown together, so 36 outcomes. Verified: a 5 appearing neither time is favoured by 5 × 5 = 25 of them, giving 25/36; a 5 appearing at least once is the denial, giving 1 − 25/36 = 11/36. The cleanest "at least" question in the exercise.

Figures to have open

• A split outcome bar: one strip holding every outcome, cut once, the two pieces labelled as the event and its denial and shown to reassemble. This carries sections 2, 4 and 5 and the chapter prints nothing like it. Must be built, and it should be the visual signature of this topic.
• The same bar with a third slice cut out of it, to show the tennis-match assumption failing when a draw is possible. Must be built; it is the picture for the one caveat in this brief.
• A four-cell panel for the two coins — HH, HT, TH, TT — with three shaded and one left over. Standard schematic, shared with Coins, dice, bags and a deck of 52: getting the denominator right.
• A 36-cell grid with the row and column for a 5 struck out, leaving a 5 × 5 block standing, for question 24. Must be built; it is what makes 25 out of 36 obvious rather than asserted.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 14 "Probability", §14.1, p. 206 — the two patterns, the notation, and the rule
• Example 4 and its remark, pp. 207–208
• Examples 5 and 6, p. 208; the note closing Example 7, p. 209; the note closing Example 9, p. 210
• Exercise 14.1 questions 1(i), 5, 7, 8, 21 and 24, pp. 214–216
• §14.2 Summary, p. 217, point 6
• Backward pointer: the totalling argument is Single-outcome events, and why all of them together come to 1; the four-outcome coin model belongs to Coins, dice, bags and a deck of 52: getting the denominator right