Lesson 12: Probability

Probability begins in Class IX as an idea — favourable outcomes over total outcomes — and Class X states it cleanly for equally likely outcomes: P(E) = (number of outcomes favourable to E) / (total number of possible outcomes). Every problem is counting, done carefully. The sample spaces are small and standard: coins, dice, cards, balls in a bag, and the calendar.

Ground rules

• 0 ≤ P(E) ≤ 1 always; P(sure event) = 1, P(impossible event) = 0.
• P(not E) = 1 − P(E) — often the fastest route ("at least one" problems!).
• A standard dice throw has 6 outcomes; two dice have 36; a coin tossed twice has 4 (HH, HT, TH, TT — order matters).

Worked example 1 (basic — dice). A die is thrown once. P(getting a number greater than 4): favourable outcomes {5, 6} → 2/6 = 1/3. P(not getting a 6) = 1 − 1/6 = 5/6.

Worked example 2 (intermediate — cards). One card is drawn from a well-shuffled pack of 52. (i) P(red king): there are 2 red kings → 2/52 = 1/26. (ii) P(face card): jack, queen, king of 4 suits = 12 → 12/52 = 3/13. (iii) P(ace of spades) = 1/52. The pack's anatomy — 4 suits, 13 ranks, 26 red and 26 black, 12 face cards, 4 aces — must be recalled instantly; fumbling it costs more time than the probability itself.

Worked example 3 (NCERT exemplar style — two dice). Two dice are thrown together. Find P(sum greater than 9) and P(doublet). Sums of 10, 11, 12 occur in 3 + 2 + 1 = 6 outcomes — (4,6), (5,5), (6,4); (5,6), (6,5); (6,6) — so P(sum > 9) = 6/36 = 1/6. Doublets: (1,1)…(6,6) = 6 outcomes → P = 6/36 = 1/6. For two-dice sums, the counts run 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums 2 through 12 — sums near 7 are the most likely, the extremes the least.

Competency application: quality control on the factory floor

A box holds 5 red and 3 blue pens, indistinguishable except by colour. A checker draws one pen at random. P(blue) = 3/8; P(not red) = 1 − 5/8 = 3/8 — the same answer by the complement, because "not red" and "blue" coincide here. Now scale it up, as case-based sets do: if 2 pens are drawn one after another *without replacement*, P(both red) = (5/8) × (4/7) = 20/56 = 5/14 — the second denominator drops to 7 and the red count to 4, which is the entire idea of conditional counting at school level.

Common mistakes

• Counting (HT) and (TH) as one outcome for two coins but then treating the sample space as 3 outcomes with equal weights. Either list 4 ordered outcomes (each 1/4) or weight the merged outcome 2/4 — never mix.
• For dice sums, assuming all sums are equally likely. Sum 7 has 6 outcomes; sum 2 has 1. Probabilities of sums are *not* uniform.
• Forgetting to reduce the fraction — CBSE expects simplest form: 12/52 must be reported as 3/13.
• Using P(E) = favourable/total when outcomes are not equally likely (a biased die, a loaded spinner). The formula carries an assumption; state it.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. If P(E) = 0.05, then P(not E) equals: (a) 0.95 (b) 0.05 (c) 1.05 (d) 0.5. Answer: (a) 0.95. Probabilities of an event and its complement sum to 1.

Q2. A die is thrown twice. The probability that the sum is 5 is: (a) 1/6 (b) 1/9 (c) 5/36 (d) 1/12. Answer: (b) 1/9. Outcomes: (1,4), (2,3), (3,2), (4,1) — 4 of 36 = 1/9.

Section B — Short answer (3 marks) Q3. A bag contains 5 red and 3 blue balls. A ball is drawn at random. Find the probability that it is (i) blue (ii) not red. Marking scheme: 1 mark for total outcomes = 8; 1 mark for (i) 3/8; 1 mark for (ii) 1 − 5/8 = 3/8 (or direct count of non-red balls = 3, equally acceptable).

Section C — Case-based long answer (5 marks) Q4. A card is drawn at random from a standard pack of 52 cards. Find the probability that the card is: (i) a king (1 mark) (ii) a red card (1 mark) (iii) the ace of spades (1 mark) (iv) not a face card (2 marks) Marking scheme: (i) 4/52 = 1/13 (1 mark). (ii) 26/52 = 1/2 (1 mark). (iii) 1/52 (1 mark). (iv) Face cards number 12, so non-face cards number 40 (1 mark); P = 40/52 = 10/13 (1 mark — also acceptable via 1 − 3/13).

Going deeper: the two-dice sum table

All 36 outcomes of two dice, organised by sum, give the counts 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 for sums 2–12. Harvest it: P(sum is prime) — prime sums are 2, 3, 5, 7, 11 with counts 1 + 2 + 4 + 6 + 2 = 15, so P = 15/36 = 5/12. P(sum is even) = (1 + 3 + 5 + 5 + 3 + 1)/36 = 18/36 = 1/2 — as symmetry demands. Building this table once and reusing it is faster than recounting outcomes in every question, and it exposes the chapter's core insight visibly: sums are *not* equally likely, outcomes are.

Going deeper: three coins and "at least"

A coin tossed three times has 8 equally likely outcomes. P(exactly two heads): the outcomes HHT, HTH, THH give 3/8. P(at least two heads): add HHH → 4/8 = 1/2. P(at least one head) = 1 − P(TTT) = 1 − 1/8 = 7/8 — the complement rule again, now doing real work: counting "at least one" directly means three separate counts, while the complement is a single outcome. When the favourable set is large and the unfavourable set is tiny, always attack from behind.

Going deeper: the calendar — probability in disguise

A leap year has 366 days = 52 full weeks + 2 extra days. The two extra days are equally likely to be any of the 7 consecutive pairs: (Sat, Sun), (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat). A 53rd Sunday occurs exactly when Sunday is among the extra pair: pairs (Sat, Sun) and (Sun, Mon) — 2 favourable of 7, so P = 2/7. For an ordinary year (1 extra day), the same question gives 1/7. This NCERT problem is pure sample-space thinking wearing a calendar costume; the "equally likely" objects are day-pairs, not days.

Going deeper: a game spinner

A spinner numbered 1 to 8 (equal sectors) is spun once. P(prime number): primes are 2, 3, 5, 7 → 4/8 = 1/2. P(multiple of 3): 3, 6 → 2/8 = 1/4. P(prime *or* multiple of 3): the sets are disjoint here (no number is both), so add: 1/2 + 1/4 = 3/4. When sets overlap — say prime or even, sharing the number 2 — the overlap must be subtracted once: P = 4/8 + 4/8 − 1/8 = 7/8. That "add, then subtract the double-counted" idea is the inclusion–exclusion principle in embryo.

Boundary cases: probability 1 and probability 0

The scale has meaningful endpoints. P(sum ≤ 12 with two dice) = 36/36 = 1: a *certain* event — every outcome qualifies. P(sum = 13) = 0/36 = 0: an *impossible* event. Every genuine uncertainty lives strictly between. A subtler "at least" computation: P(at least one six in two dice) = 1 − P(no six) = 1 − 25/36 = 11/36 (the no-six outcomes are 5 × 5 = 25). Checking against direct counting: sixes appear in 6 + 6 − 1 = 11 outcomes (the (6,6) counted in both rows) ✓ — two routes, one answer, exactly the cross-checking habit this chapter rewards. And a vocabulary point examiners test: two events are mutually exclusive when they cannot occur together (P of both = 0, like "sum 7" and "sum 8" on one throw), and equally likely when symmetry gives them the same probability. Every computation in this chapter quietly assumes the second and often exploits the first — naming them correctly is worth the one-mark theory questions that surround the numericals.

Final drills

Drill 1 — dice differences. Two dice are thrown. Find P(the absolute difference of the numbers is 1). The favourable outcomes: (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5) — ten of them (each adjacent pair, in both orders). P = 10/36 = 5/18. Counting in *ordered* pairs keeps every outcome at weight 1/36; unordered counting tempts double-counting errors for doubles and undercounting for non-doubles.

Drill 2 — compound card questions. A card is drawn from a pack of 52. (i) P(king or queen): the events are disjoint, 8 favourable cards, P = 8/52 = 2/13. (ii) P(black face card): 3 face ranks × 2 black suits = 6 cards, P = 6/52 = 3/26. (iii) P(red *and* king): 2 cards, P = 1/26 — "or" adds (minus any overlap), "and" restricts. Reading the connective correctly is worth more marks in this chapter than any formula.

Drill 3 — a fair-game intuition. A quiz has four options per question; a student guesses all 20 questions blindly. On average she expects 20 × 1/4 = 5 correct — probability as long-run frequency. This "expected count = n × p" idea is not in the Class X syllabus formally, but it explains why P is defined as a ratio at all: favourable/total is the fraction of trials that succeed in the long run.