- Describe the sample space of an experiment and compute P = m/n when the outcomes are equally likely.
- Count outcomes by listing, with the multiplication principle, and with permutations and combinations.
- Use complementary, mutually exclusive and independent events, including “at least one” questions.
- Solve basic geometric-probability problems with lengths and areas.
Toss two coins. Are “two heads”, “two tails” and “one head and one tail” equally likely, 1/3 each? Many people say yes, and they are wrong: “one head and one tail” happens in two ways, HT and TH, so its probability is 1/2, while each of the other two has probability 1/4. Classical probability (古典概型) is the skill of listing equally likely outcomes correctly and counting them fast. The CSCA syllabus names it first in the “Probability and statistics” part: “classical probability model and probability calculation”. The other two lines of that part come in the next lessons, “Numerical characteristics of data” and “The normal distribution”; the set operations that describe events are in “Sets and set operations”.
Sample space, events and the classical model
The sample space Ω (样本空间) is the set of all possible outcomes of a random experiment; each outcome is a sample point. An event (随机事件) is a subset of Ω. Ω itself is the certain event, and ∅ is the impossible event. Rolling a die: Ω = {1, 2, 3, 4, 5, 6}, and “an even number comes up” is the event A = {2, 4, 6}.
To use the formula below, the outcomes must be equally likely (等可能). That is why we label objects that look the same: two coins give Ω = {HH, HT, TH, TT}, four equally likely outcomes, not three; two dice give 6 · 6 = 36 ordered pairs.
An experiment is a classical probability model (古典概型) if it has finitely many outcomes and all of them are equally likely. Rolling a fair die is one; shooting at a target is not (the rings are not equally likely to be hit), and neither is choosing a real number from an interval (there are infinitely many outcomes; see the last section).
- nnumber of all equally likely outcomes (the size of Ω)
- mnumber of outcomes favourable to A
- P(A)probability of A (概率)
First make the outcomes equally likely, then count; m and n must be counted in the same way (both ordered or both unordered).
1) A die is rolled. Find the probability of a number greater than 4, and of a prime number.
2) Three coins are tossed. Find the probability of exactly two heads.
3) Two dice are rolled. Find the probability that the sum is 9, and that the two numbers are equal.
Show solutionHide solution
2) Ω has 2³ = 8 outcomes; exactly two heads: HHT, HTH, THH, so P = 3/8.
3) Sum 9: (3, 6), (4, 5), (5, 4), (6, 3), P = 4/36 = 1/9. Equal numbers: (1, 1), …, (6, 6), P = 6/36 = 1/6.
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Why do we trust P = m/n? Repeat the experiment many times: the relative frequency of an event (频率, the number of times it happened divided by the number of trials) settles down near a fixed number, and for equally likely outcomes that number is m/n. The classical formula predicts the long-run frequency; it promises nothing about the next few trials. Six rolls of a die do not have to show a six.
Three letters are put at random into three addressed envelopes, one letter in each envelope. What is the probability that exactly one letter is in its correct envelope?
A) 1/3 B) 1/6 C) 1/2 D) 2/3
Show solutionHide solution
A is the probability that no letter is in its envelope (231, 312). B counts only one arrangement (for example 132) and forgets that any of the three letters can be the correct one. D also adds the case where all three are correct.
Counting: the multiplication principle, permutations and combinations
Listing works for up to a few dozen outcomes. Beyond that, count with two principles. Addition principle (分类加法): if a task can be done in one of several separate ways (classes), the numbers of ways add. Multiplication principle (分步乘法): if a task is done in several steps one after another, and the number of choices at each step does not depend on the earlier choices, the numbers multiply.
- n₁, n₂, …, nₖnumber of ways in each class or at each step
- Ntotal number of ways
Ask yourself: is the task finished after one choice (add) or only after all the steps (multiply)?
1) There are 3 roads from A to B, 2 roads from B to C and 2 direct roads from A to C. How many routes lead from A to C?
2) How many three-digit numbers can be written with the digits 1, 2, 3, 4, 5 if digits may repeat? If they may not?
3) In how many ways can 4 letters be posted in 3 mailboxes?
Show solutionHide solution
2) With repetition 5 · 5 · 5 = 125; without repetition 5 · 4 · 3 = 60.
3) Each letter chooses one of the 3 boxes independently: 3 · 3 · 3 · 3 = 3⁴ = 81. Not 4³ = 64: the boxes do not choose letters.
Take the letters a, b, c and choose two of them. If order matters (a president and a secretary), there are 3 · 2 = 6 results: ab, ba, ac, ca, bc, cb. If it does not (a pair on duty), ab and ba are the same pair, so only 6/2 = 3 remain: {a, b}, {a, c}, {b, c}. Dividing by the number of orders inside each group is exactly how combinations come from permutations.
- n!1 · 2 · … · n (0! = 1)
- Aₙᵏarrangements (permutations, 排列) of k objects chosen from n: order matters
- Cₙᵏcombinations (组合) of k objects chosen from n: order does not matter
Each combination of k objects can be ordered in k! ways, so Aₙᵏ = k! · Cₙᵏ. Chinese textbooks use the same symbols Aₙᵏ and Cₙᵏ; some books write P(n, k) and C(n, k).
1) In how many ways can 4 different books stand on a shelf? How many handshakes are there when 8 people all shake hands with each other? In how many ways can 8 people choose a president and a treasurer?
2) A team of 2 boys and 1 girl is chosen from 5 boys and 4 girls. How many different teams are possible?
3) A bag holds 4 red and 3 blue balls, and two balls are drawn at the same time. Find the probability that both are red, and that the colours are different.
Show solutionHide solution
2) Two steps: C₅² · C₄¹ = 10 · 4 = 40.
3) n = C₇² = 21. Both red: C₄² = 6, P = 6/21 = 2/7. Different colours: 4 · 3 = 12, P = 12/21 = 4/7. Check: blue–blue C₃² = 3, and 6 + 12 + 3 = 21.
3 students are chosen at random from 4 boys and 3 girls. What is the probability that exactly 2 of the chosen students are boys?
A) 12/35 B) 18/35 C) 22/35 D) 6/35
Show solutionHide solution
A is the probability of exactly one boy (C₄¹ · C₃² = 12). C is “at least two boys” (18 + 4 = 22). D counts only the pairs of boys (C₄² = 6) and forgets to choose the girl.
Complementary, mutually exclusive and independent events
A and B are mutually exclusive (互斥事件) if they cannot happen together: A ∩ B = ∅. They are complementary (对立事件) if, in addition, one of them must happen: A ∪ B = Ω; then B = Ā. Rolling a die: “1” and “6” are mutually exclusive but not complementary; “even” and “odd” are complementary.
- Āthe complement of A (“A does not happen”)
- A ∪ BA or B (at least one of them)
- AB = A ∩ BA and B together
The last formula is inclusion–exclusion from “Sets and set operations”: the common part AB would otherwise be counted twice. Complementary events are always mutually exclusive, but not the other way round.
1) A die is rolled. A = “1 or 2”, B = “6”, C = “an even number”, D = “a number greater than 3”. Find P(A ∪ B) and P(C ∪ D).
2) A die is rolled twice. Find the probability of at least one six.
3) P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8. Find P(AB). Are A and B mutually exclusive?
Show solutionHide solution
2) Complement: no six in either roll, 5 · 5 = 25 of the 36 outcomes. P = 1 − 25/36 = 11/36.
3) P(AB) = 0.6 + 0.5 − 0.8 = 0.3 ≠ 0, so they are not mutually exclusive.
A and B are independent (相互独立事件) if the occurrence of one does not change the probability of the other. Different shooters, different machines and draws with replacement (有放回) give independent events; draws without replacement (不放回) do not, because the first draw changes the contents of the bag.
Independence is a statement about numbers, not about devices: A and B are independent exactly when P(AB) = P(A) · P(B). Even one roll of a die can give independent events: for A = “even” and B = “1 or 2”, P(AB) = P({2}) = 1/6 = 1/2 · 1/3. In exam problems, however, independence usually comes from the situation (different people, separate machines, draws with replacement), and you simply use the product rule.
- p₁, …, pₙprobabilities of independent events
- 1 − pᵢprobability that the i-th event does not happen
If A and B are independent, so are Ā and B, A and B̄, and Ā and B̄. Exactly one of two: P = p₁(1 − p₂) + (1 − p₁)p₂.
1) Two machines work independently; the probabilities that they work all day without a fault are 0.9 and 0.8. Find the probabilities that both work, that neither works, that exactly one works and that at least one works.
2) A bag holds 4 white and 2 black balls. Two balls are drawn one after the other. Find the probability that both are black if the first ball is put back, and if it is not.
3) A shooter hits a target with probability 0.5 per shot, independently. What is the smallest number of shots for which the probability of at least one hit exceeds 0.9?
Show solutionHide solution
2) With replacement: (2/6) · (2/6) = 1/9. Without: (2/6) · (1/5) = 1/15, the same as C₂²/C₆² = 1/15.
3) 1 − 0.5ⁿ > 0.9 ⇔ 0.5ⁿ < 0.1: 0.5³ = 0.125 is too big, 0.5⁴ = 0.0625 works, so n = 4.
Two coins are tossed. Which pair of events is mutually exclusive but NOT complementary?
A) “at least one head” and “at least one tail”
B) “at least one head” and “two tails”
C) “two heads” and “at most one head”
D) “exactly one head” and “two heads”
Show solutionHide solution
A: HT belongs to both, so they are not even mutually exclusive. B and C are complementary pairs (together they cover Ω), which is more than the question asks for.
Geometric probability
If the outcomes are the points of a segment, a region or a solid, there are infinitely many of them and P = m/n cannot be used. When the point is chosen uniformly at random (pieces of equal size are equally likely), the probability of an event is the share of the length, area or volume it occupies: the geometric probability model (几何概型).
- Ωthe whole segment, region or solid
- Athe part where the event happens
A single point has length 0, so its probability is 0; that is why “<” and “≤” give the same answer here.
1) A number x is chosen at random from [0, 5]. Find the probability that x² − 3x + 2 < 0.
2) A bus comes every 12 minutes, and you arrive at the stop at a random moment. Find the probability that you wait at most 3 minutes.
3) A point is chosen at random in a square with side 2. Find the probability that it is less than 1 away from the centre, and that it is less than 1 away from a given vertex.
Show solutionHide solution
2) Favourable: the last 3 minutes before a bus, 3 out of 12, P = 1/4.
3) The disc of radius 1 around the centre lies inside the square: P = π · 1²/2² = π/4. Around a vertex only a quarter of the disc lies in the square: P = (π/4)/4 = π/16.
Two random numbers x and y, each chosen from an interval, form one random point (x, y) of a rectangle, so questions about two random moments (two friends arriving at a café, two buses) become area questions. Draw the rectangle, shade the region where the condition holds and compare the areas; the region is usually bounded by straight lines such as x + y = 1 or y = x.
A point (x, y) is chosen at random in the square with vertices (0, 0), (2, 0), (2, 2) and (0, 2). What is the probability that x + y ≥ 1?
A) 7/8 B) 1/8 C) 3/4 D) 1/2
Show solutionHide solution
B is the probability of the complement, x + y < 1. C takes the area of the triangle as 1. D uses the diagonal x + y = 2, which halves the square.
How CSCA asks about this
Probability items in the CSCA are short word problems about dice, coins, balls, cards, people in a row, shooters or random points. Learn the key terms in English and Chinese:
| Term | 中文 | Pinyin |
|---|---|---|
| probability | 概率 | gàilǜ |
| sample space | 样本空间 | yàngběn kōngjiān |
| random event | 随机事件 | suíjī shìjiàn |
| classical probability model | 古典概型 | gǔdiǎn gàixíng |
| equally likely | 等可能 | děng kěnéng |
| addition principle | 分类加法计数原理 | fēnlèi jiāfǎ jìshù yuánlǐ |
| multiplication principle | 分步乘法计数原理 | fēnbù chéngfǎ jìshù yuánlǐ |
| permutation (arrangement) | 排列 | páiliè |
| combination | 组合 | zǔhé |
| mutually exclusive events | 互斥事件 | hùchì shìjiàn |
| complementary events | 对立事件 | duìlì shìjiàn |
| independent events | 相互独立事件 | xiānghù dúlì shìjiàn |
| geometric probability model | 几何概型 | jǐhé gàixíng |
| with / without replacement | 有放回 / 不放回 | yǒu fànghuí / bù fànghuí |
Typical question patterns and time-savers (about 75 s per item):
- Dice, coins, balls: label identical objects so that the outcomes are equally likely; for two dice use the 6 × 6 table (n = 36).
- Several objects drawn at once: combinations in both m and n; one after another: the multiplication principle in both. Never mix the two.
- “At least one”: 1 − P(none); “exactly one of two”: p₁(1 − p₂) + (1 − p₁)p₂.
- Statements about events: mutually exclusive means no common outcome; complementary means no common outcome and nothing left over; independent means P(AB) = P(A)P(B).
- Arrangements: “together” means a block; “not together” means gaps; in digit problems handle the restricted place first (no 0 in front).
- Geometric probability: draw the segment or region and compare lengths or areas; the boundary does not matter.
Key points
- Classical model: finitely many equally likely outcomes, P(A) = m/n; make the outcomes equally likely first (label coins, dice and balls).
- Multiply for steps one after another, add for separate cases; Aₙᵏ = n!/(n − k)! when order matters, Cₙᵏ = n!/(k!(n − k)!) when it does not.
- P(Ā) = 1 − P(A); “at least one” = 1 − P(none).
- Mutually exclusive: P(A ∪ B) = P(A) + P(B); in general P(A ∪ B) = P(A) + P(B) − P(AB); independent: P(AB) = P(A)P(B).
- Geometric probability is the ratio of the lengths, areas or volumes of the favourable region and of the whole region.
Check yourself
12 questions. Every correct answer earns XP.
Topic test: 20 questions · 25 min
Finished the lesson? Check yourself with a timed test on this topic.