Unit 11 · Counting and Probability　计数与概率

By the end of this chapter you can:

1. Apply the addition and multiplication principles to count outcomes in multi-step tasks
2. Compute permutations $P(n,k)$ and combinations $\binom{n}{k}$ and know when to use each
3. Calculate probabilities in classical (equally-likely) models using favorable / total outcomes
4. Use independence and the addition rule — including the inclusion-exclusion correction — to find probabilities of compound events

Exam weight on past CSCA papers: ~6% (3 of 48 MCQs).

11.1　Counting Principles　计数原理

The two fundamental rules

Addition Principle (加法原理): If task $A$ can be done in $m$ ways and task $B$ can be done in $n$ ways, and the two tasks are mutually exclusive (doing $A$ prevents doing $B$), then either $A$ or $B$ can be done in $m + n$ ways.

Multiplication Principle (乘法原理): If a multi-step procedure has step 1 done in $m$ ways and step 2 done in $n$ ways (regardless of what step 1 chose), then the complete procedure can be done in $m \times n$ ways.

🔑 Key question to distinguish them: Does your final result require one action (use addition) or does it require several sequential actions (use multiplication)?

Worked Example 11.1.A

A school offers 4 science electives and 3 humanities electives. A student must choose either one science or one humanities elective for a free period. In how many ways can they choose?

Solution. The student picks from science or humanities — these are mutually exclusive categories. Use the Addition Principle:

$$4 + 3 = \boxed{7}$$

Contrast: If the student instead had to pick one science and one humanities elective, the answer would be $4 \times 3 = 12$ (Multiplication Principle).

⚠️ Trap — "or" does not always mean add: The addition principle requires truly mutually exclusive options. If some choices belong to both categories simultaneously, you must subtract the overlap (see 11.5 for inclusion-exclusion). Only add directly when the categories cannot overlap.

The multiplication principle across multiple steps

For a procedure with $k$ independent steps with $n_1, n_2, \ldots, n_k$ choices respectively:

$$\text{Total outcomes} = n_1 \times n_2 \times \cdots \times n_k$$

Example: A password is 3 letters (A–Z) followed by 2 digits (0–9), no restrictions.

$$\text{Total passwords} = 26^3 \times 10^2 = 17{,}576 \times 100 = 1{,}757{,}600$$

11.2　Permutations and Combinations　排列与组合

Permutations — order matters　排列（顺序有意义）

A permutation is an ordered selection of $k$ objects from $n$ distinct objects.

$$P(n,k) = \frac{n!}{(n-k)!} = n(n-1)(n-2)\cdots(n-k+1)$$

Reading: "How many ways to arrange $k$ items chosen from $n$?" Fill $k$ slots: the first has $n$ choices, the second has $n-1$, …, the $k$-th has $n-k+1$.

Special cases:

• $P(n,n) = n!$ — all $n$ objects arranged in a line
• $P(n,1) = n$

Combinations — order does not matter　组合（顺序无意义）

A combination is an unordered selection of $k$ objects from $n$ distinct objects.

$$\binom{n}{k} = \frac{n!}{k!\,(n-k)!} = \frac{P(n,k)}{k!}$$

The factor $k!$ divides out the overcount from all $k!$ orderings of the same chosen $k$ objects.

Pascal-like identities (useful for verification)

$$\binom{n}{k} = \binom{n}{n-k} \qquad \text{(choosing $k$ to include = choosing $n-k$ to exclude)}$$

$$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} \qquad \text{(Pascal's Identity)}$$

$$\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n$$

🔑 Permutation vs. Combination decision rule:

• Order matters (rank, row seats, passwords, race positions) → $P(n,k)$
• Order does not matter (committee, hand of cards, sample, group) → $\binom{n}{k}$

Worked Example 11.2.A

A team of 10 players must elect a captain and a vice-captain (two different roles). Then separately, a 3-player sub-group is chosen from the remaining 8 for a special task. How many ways can this be done?

Solution.

Step 1 — Choose captain and vice-captain (order matters — the roles are distinct):

$$P(10,2) = 10 \times 9 = 90$$

Step 2 — Choose 3 players from the remaining 8 (order does not matter — it is just a group):

$$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$

Step 3 — Multiply (two independent sequential steps):

$$90 \times 56 = \boxed{5040}$$

⚠️ Trap — counting with restrictions: When a problem says "must include person X" or "cannot place A next to B," handle the restriction first (fix what is forced), then count the remaining free choices. Do not apply the formula to the full pool and subtract after, unless the subtraction is straightforward.

11.3　Classical Probability　古典概型

Definition

A classical probability model (古典概型) has a finite sample space $\Omega$ where every outcome is equally likely. The probability of event $A$ is:

$$P(A) = \frac{\text{number of outcomes in } A}{\text{total number of outcomes in } \Omega} = \frac{|A|}{|\Omega|}$$

Basic properties:

Worked Example 11.3.A

A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both balls are red?

Solution.

Total ways to choose 2 from 8 (unordered — any pair is equally likely):

$$|\Omega| = \binom{8}{2} = 28$$

Favorable outcomes (both red — choose 2 from the 5 red balls):

$$|A| = \binom{5}{2} = 10$$

$$P(A) = \frac{10}{28} = \boxed{\dfrac{5}{14}}$$

🔑 Why combinations, not permutations? When drawing without replacement and asking for a group (not "which one drawn first"), the draw is unordered — use $\binom{n}{k}$ for both numerator and denominator. The equally-likely assumption holds because each unordered pair is equally likely to be selected.

11.4　Independent Events　独立事件

Definition

Two events $A$ and $B$ are independent (独立) if knowing that $A$ occurred gives no information about whether $B$ occurs:

$$P(A \cap B) = P(A) \cdot P(B)$$

This is both the definition and the computational rule. If the equation holds, the events are independent; if it fails, they are dependent.

Extension to three or more events: $A$, $B$, $C$ are mutually independent if

$$P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$$

(all pairwise independence conditions must also hold separately).

With replacement vs. without replacement

🔑 The with/without replacement distinction is one of the most commonly tested traps on CSCA. Always check whether the problem says 放回 or 不放回 before writing down any probability formula.

Worked Example 11.4.A

A fair coin is flipped and a fair six-sided die is rolled. What is the probability of getting heads and a 4?

Solution. The coin and die are physically independent.

$$P(\text{heads}) = \frac{1}{2}, \qquad P(\text{rolling a 4}) = \frac{1}{6}$$

$$P(\text{heads and 4}) = \frac{1}{2} \times \frac{1}{6} = \boxed{\dfrac{1}{12}}$$

Repeated independent trials — "at least once" shortcut

If the probability of success in one trial is $p$, and $n$ independent trials are conducted:

$$P(\text{at least one success}) = 1 - P(\text{zero successes}) = 1 - (1-p)^{n}$$

⚠️ Trap — "at least once" problems: Students often try to add $P(\text{exactly 1}) + P(\text{exactly 2}) + \cdots$, which is tedious and error-prone. Always use the complement: $1 - P(\text{none})$.

11.5　Mutually Exclusive Events and the Addition Rule　互斥事件与加法公式

General addition rule (inclusive form)

For any two events $A$ and $B$:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

The term $P(A \cap B)$ corrects for outcomes counted in both $A$ and $B$ (inclusion-exclusion).

Mutually exclusive events (互斥事件)

Events $A$ and $B$ are mutually exclusive (also called disjoint, 互斥) if they cannot both occur:

$$A \cap B = \emptyset \implies P(A \cap B) = 0$$

The addition rule then simplifies to:

$$P(A \cup B) = P(A) + P(B)$$

For a collection of pairwise mutually exclusive events $A_1, A_2, \ldots, A_n$:

$$P(A_1 \cup A_2 \cup \cdots \cup A_n) = P(A_1) + P(A_2) + \cdots + P(A_n)$$

Complementary events (对立事件) — a special case

$\overline{A}$ (the complement of $A$) is always mutually exclusive with $A$, and together they exhaust $\Omega$:

$$P(A) + P(\overline{A}) = 1 \qquad \Longleftrightarrow \qquad P(\overline{A}) = 1 - P(A)$$

⚠️ Trap — forgetting the overlap in the general addition rule: A common error is writing $P(A \cup B) = P(A) + P(B)$ when $A$ and $B$ are not mutually exclusive. This overcounts the intersection. Always check whether $A \cap B = \emptyset$ before simplifying.

Worked Example 11.5.A

A card is drawn at random from a standard 52-card deck. Find the probability that the card is a king or a heart.

Solution. Let $A$ = "king", $B$ = "heart."

$$P(A) = \frac{4}{52}, \quad P(B) = \frac{13}{52}, \quad P(A \cap B) = \frac{1}{52} \text{ (the king of hearts)}$$

These events are not mutually exclusive (the king of hearts is both). Apply the general rule:

$$P(A \cup B) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \boxed{\dfrac{4}{13}}$$

Try it!　自测练习

Q1. A student can travel to school by bus (3 routes) or by metro (2 lines). How many total travel options are there?

Q2. From 6 students, how many ways can a president, vice-president, and secretary be elected (three distinct roles)?

Q3. A committee of 4 is chosen at random from 9 people. What is the probability that a specific person, Alice, is included?

Q4. A fair die is rolled twice (independent rolls). What is the probability of getting a number greater than 4 on both rolls?

Q5. $P(A) = 0.4$, $P(B) = 0.3$, $P(A \cap B) = 0.1$. Find $P(A \cup B)$.

📌 Chapter summary

Topic	Key formula / rule
11.1 Counting principles	Mutually exclusive options → add; sequential steps → multiply
11.2 Permutations	$P(n,k) = \dfrac{n!}{(n-k)!}$ — order matters
11.2 Combinations	$\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ — order does not matter
11.2 Pascal identity	$\binom{n}{k} = \binom{n}{n-k}$; sum of row $= 2^n$
11.3 Classical probability	$P(A) = |A|/|\Omega|$ — equally likely outcomes only
11.3 Complement	$P(\overline{A}) = 1 - P(A)$ — shortcut for "at least one"
11.4 Independence	$P(A \cap B) = P(A) \cdot P(B)$; with replacement → independent
11.5 Addition rule	$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
11.5 Mutually exclusive	$A \cap B = \emptyset$ → $P(A \cup B) = P(A) + P(B)$

What's next → Unit 12 covers Statistics: mean, variance, and frequency distributions — where probability models connect to describing real data sets.