PrepShorts · Teaching notes · Class 10 Mathematics · Chapter 14, ProbabilityPrepShorts

Chapter 14 · Probability

Coins, dice, bags and a deck of 52: getting the denominator right

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16 min.

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What the video covers, what to say before it, where a class usually goes wrong, and what to set afterwards. An account is free, and it opens every chapter of every book.

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What to assume they know

What they should be able to do

  • Fix the denominator for a described experiment by listing or counting outcomes that are interchangeable, before computing anything
  • Total the contents of a bag, box, bank or lot to obtain that denominator
  • Describe the structure of a 52-card pack — suits, colours, ranks and face cards — and use it to produce a numerator for any described card
  • Build the four-outcome list for two distinguishable coins and explain why three descriptions are not three outcomes
  • Build the 36-cell grid for two dice and read a numerator off it
  • Explain why an ordered pair and its reverse are two outcomes and not one
  • Explain why the eleven possible totals on two dice are not equally likely
  • Adjust the denominator for a second draw made without replacement
  • Recognise the recurring board phrasings — at random, well-shuffled, identical cards — as guarantees that the outcomes are interchangeable

Where it usually goes wrong

  • "Two coins give three results." They give four. The chapter spends a whole example and an exercise question on this, and gives the coins different values so the student cannot claim the two mixed results are the same one.
  • "Eleven possible totals, so 1/11 each." Question 22(ii) is built to draw this out. The totals do cover everything without overlap, which is exactly why it feels safe; what fails is that they are not the same size.
  • "(1,4) and (4,1) are the same throw." The chapter asks the reader why they differ and does not answer. The dice are different colours, so the two are physically distinguishable arrangements.
  • "Face cards include the aces." They do not — the chapter lists kings, queens and jacks, which is 12 cards, not 16. This costs marks on question 14 every year.
  • "Taking a card out and not replacing it changes nothing." Questions 15 and 17 both drop the denominator by one between their parts, and question 15's second part drops a numerator to zero as well.
  • "The denominator is the last number the question mentions." In question 10 no printed number is the denominator at all — it has to be assembled from four of them. In question 16 the same is true.
  • "A number lying between 2 and 6 includes 2 and 6." Question 13 is read strictly, giving 3, 4 and 5. Say so out loud, because the reading is the whole question.
  • "Different-coloured dice is a detail of the story." It is the argument. Grey and blue is what makes 36 rather than 21 the honest count.

Questions to check understanding

  • Compute the denominator from a described container and then answer two or three parts off it — by far the commonest form in this exercise
  • Card questions across suit, colour, rank and face-card descriptions, as in question 14
  • Complete a probability table for the totals of two dice, as in question 22(i)
  • Judge whether a supplied counting argument is sound, as in questions 22(ii) and 25(i)
  • Two-part questions where the second draw follows the first without replacement, as in questions 15 and 17
  • List all outcomes of a small compound experiment before computing, which is what Example 13 asks for in its own opening

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data; the chapter prints answers only for its worked examples, never for the exercise.

  • The pack (p. 207). 52 cards; four suits of 13; the suit symbols for spades, hearts, diamonds and clubs are printed inline and are read off the printed page. Clubs and spades black, hearts and diamonds red. Each suit runs ace, king, queen, jack, then 10 down to 2. Face cards are the kings, queens and jacks. The thirteen-per-suit count is stated on the page itself. Verified consequences the chapter does not print: 26 cards of each colour, 12 face cards, 6 red face cards, and 4 of any given rank.
  • Example 4 (pp. 207–208). Four aces, so 4/52 = 1/13; not an ace, 48/52 = 12/13.
  • Example 7 (p. 209). Forty students, 25 girls and 15 boys, one name written on each of forty identical cards. Verified: 25/40 = 5/8 and 15/40 = 3/8. The cards being identical is the guarantee, not scene-setting.
  • Example 8 (p. 209). 3 blue, 2 white, 4 red marbles. Verified: the denominator is 9, and the three answers are 2/9, 1/3 and 4/9.
  • Example 9 (p. 210). Two coins tossed together, deliberately of different values so they can be told apart — one of ₹1 and one of ₹2. The four outcomes are HH, HT, TH, TT. Verified: at least one head is favoured by three of the four, giving 3/4.
  • Example 12 (p. 212). A carton of 100 shirts: 88 good, 8 with minor defects, 4 with major defects. One trader accepts only the good ones, the other rejects only the majors. Verified: 88/100 = 0.88 and (88 + 8)/100 = 0.96.
  • Example 13 and Fig. 14.3 (pp. 212–213). Two dice thrown together, made distinguishable by colour — one blue, the other grey. The grid lists all 36 ordered pairs, the first entry being the blue die's. Verified: 6 × 6 = 36; a total of 8 is favoured by (2,6), (3,5), (4,4), (5,3) and (6,2), so 5/36. The printed figure draws a small die above the columns and another to the left of the rows, and rings the five cells that make 8 — worth keeping in the redraw, because it shows the favourable set running diagonally rather than in a block.
  • Exercise 14.1 question 22 (p. 216). The table of totals 2 to 12, with 1/36, 5/36 and 1/36 pre-filled at 2, 8 and 12. Verified by counting Fig. 14.3 cell by cell: the counts are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, so the probabilities are 1/36, 1/18, 1/12, 1/9, 5/36, 1/6, 5/36, 1/9, 1/12, 1/18, 1/36. Part (ii) asks whether eleven totals means 1/11 apiece; verified wrong — the eleven events are a genuine covering but their sizes run from 1 to 6.
  • Exercise 14.1 question 23 (p. 216). One rupee coin tossed three times; a win requires all three the same. Verified: 8 outcomes; 2 of them win, so losing is 6/8 = 3/4.
  • Exercise 14.1 question 25(i) (p. 217). Verified: the three descriptions of two coins carry 1/4, 1/4 and 1/2, not 1/3 each. Cross-referenced from Equally likely outcomes, and the everyday cases where that fails, which owns this question; here it is the worked consequence of section 7.
  • The exercise's denominators, all verified by an added count. Question 10, a piggy bank of one hundred 50p, fifty ₹1, twenty ₹2 and ten ₹5 coins: 180 coins, so 100/180 = 5/9 and 170/180 = 17/18. Question 11, a tank of 5 male and 8 female fish drawn as Fig. 14.4: 13 fish, so 5/13. Question 12, the eight-sector spinner of Fig. 14.5: 4/8 = 1/2 for an odd number and 6/8 = 3/4 for a number above 2. Question 13, one die: primes 2, 3 and 5 give 1/2; the numbers strictly between 2 and 6 are 3, 4 and 5, giving 1/2; odd numbers give 1/2 — all three come out the same, which is a good moment. Question 14, the pack: a red king 2/52 = 1/26; a face card 12/52 = 3/13; a red face card 6/52 = 3/26; the jack of hearts 1/52; a spade 13/52 = 1/4; the queen of diamonds 1/52. Question 16, 12 defective pens mixed into 132 good ones: 144 in all, so a good one is 132/144 = 11/12. Question 18, 90 discs numbered 1 to 90: two-digit numbers are 10 to 90, so 81/90 = 9/10; perfect squares are 1, 4, 9, 16, 25, 36, 49, 64 and 81, so 9/90 = 1/10; multiples of 5 are 5 through 90, eighteen of them, so 18/90 = 1/5. Question 19, a die whose six faces read A, B, C, D, E and then a second A: an A is 2/6 = 1/3 and a D is 1/6 — the only place in the chapter where two faces carry the same label.
  • Drawing without replacement (p. 215, questions 15 and 17). Question 15: five diamonds — the ten, the jack, the queen, the king and the ace — shuffled face down. Verified: the queen is 1/5; once the queen is drawn and set aside four remain, so an ace is 1/4 and a queen is 0. Question 17: a lot of 20 bulbs with 4 defective, so a defective one is 4/20 = 1/5; if that bulb was sound and is not put back, 19 remain of which 15 are sound, so 15/19. Both questions change the denominator between their parts, and that is the entire point of including them.

Figures to have open

  • The 36-cell grid, redrawn from Fig. 14.3 (p. 213), with the two dice distinguishable at the head of the rows and columns and with individual cells able to light up. This is the workhorse figure of the topic and of question 22 and question 24; it must support shading a diagonal, a row, a column and a 5 × 5 block.
  • A histogram of the eleven totals with heights 1 to 6 and back, set beside eleven equal bars. Nothing like it is printed, and section 10 cannot be made without it. Must be built.
  • A 52-card layout, 4 rows by 13 columns, colour-coded by suit, with the face cards marked. Must be built from the description on p. 207; the chapter prints only the suit symbols inline.
  • The four-outcome coin panel, shared with Complements: knowing one probability hands you the other.
  • Fig. 14.4 (p. 215) is a small drawing of a shopkeeper at a fish tank and carries no data — the 5 and the 8 are in the question's text, not in the picture. The explanation does not need it; if it wants an image there, an icon will do.
  • Fig. 14.5 (p. 215), the eight-sector spinner, is worth rebuilding as a schematic because the sector count is the denominator.

Where this sits in the book

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