概率gàilǜ

Core Concept

Probability is a numerical measure of the likelihood that a random event will occur, describing how likely an outcome is in a random experiment.

Mathematical Definition

For a random event $A$, its probability $P(A)$ is a real number between 0 and 1:

$0 \leq P(A) \leq 1$

where:

• $P(A) = 0$: impossible event
• $P(A) = 1$: certain event
• $0 < P(A) < 1$: random event

Classical Probability

When a random experiment satisfies:

1. Finite number of possible outcomes
2. All outcomes are equally likely

Then the probability of event $A$ is:

$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{m}{n}$

Basic Properties of Probability

Property 1: Complementary Events

If events $A$ and $\overline{A}$ are complementary:

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

Property 2: Addition Rule

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

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

Special case: When $A$ and $B$ are mutually exclusive ($A \cap B = \emptyset$):

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

Property 3: Conditional Probability

The probability of event $A$ given that event $B$ has occurred:

$P(A|B) = \frac{P(A \cap B)}{P(B)} \quad (P(B) > 0)$

Property 4: Independent Events

If events $A$ and $B$ are independent:

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

Common Calculation Methods

Method 1: Enumeration

Use: When number of outcomes is small

Example: Rolling two dice, find probability sum equals 7?

Analysis:

• Total outcomes: $6 \times 6 = 36$
• Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 cases

$P = \frac{6}{36} = \frac{1}{6}$

Method 2: Combinatorics

Use: When number of outcomes is large

Example: Drawing 5 cards from 52-card deck, probability of exactly 3 Aces?

Analysis:

• Total ways: $C_{52}^5$
• Ways to get 3 Aces: $C_4^3 \times C_{48}^2$

$P = \frac{C_4^3 \times C_{48}^2}{C_{52}^5} = \frac{4 \times 1128}{2598960} \approx 0.00174$

Real-World Applications

Application 1: Lottery Problem

Problem: Box has 5 red and 3 white balls. Draw 2 balls, find probability of 1 red and 1 white.

Analysis:

• Total ways: $C_8^2 = 28$
• 1 red, 1 white: $C_5^1 \times C_3^1 = 15$

$P = \frac{15}{28}$

Application 2: Quality Control

Problem: Product has 95% pass rate. Find probability of defect when randomly selecting 1 item.

Analysis: Let $A$ = pass event, then $\overline{A}$ = defect event

$P(\overline{A}) = 1 - P(A) = 1 - 0.95 = 0.05$

Application 3: Weather Forecast

Problem: Tomorrow's rain probability is 70%. If rain, traffic jam probability is 80%; if no rain, 30%. Find jam probability.

Analysis: Let $R$ = rain, $T$ = traffic jam

$P(T) = P(R) \cdot P(T|R) + P(\overline{R}) \cdot P(T|\overline{R})$ $= 0.7 \times 0.8 + 0.3 \times 0.3 = 0.56 + 0.09 = 0.65$

Answer: 65% probability of traffic jam

CSCA Practice Problems

💡 Note: The following practice problems are designed based on the CSCA exam syllabus and Chinese standardized test formats to help students familiarize themselves with question types and problem-solving approaches.

Example 1: Basic (Difficulty ★★☆☆☆)

Randomly select one integer from 1 to 10. What's the probability of selecting an even number?

Options:

• A. $\frac{1}{10}$
• B. $\frac{1}{5}$
• C. $\frac{2}{5}$
• D. $\frac{1}{2}$

Solution:

Total outcomes: 10

Even numbers: 2, 4, 6, 8, 10 → 5 numbers

$P = \frac{5}{10} = \frac{1}{2}$

Answer: D

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Example 2: Intermediate (Difficulty ★★★☆☆)

Bag has 5 red and 3 white balls. Draw 2 balls randomly. Find probability of at least 1 red ball.

Solution:

Method 1: Direct approach At least 1 red = exactly 1 red + exactly 2 red

$P = \frac{C_5^1 \times C_3^1 + C_5^2}{C_8^2} = \frac{15 + 10}{28} = \frac{25}{28}$

Method 2: Complement (simpler!) Complement of "at least 1 red" = "0 red" (i.e., 2 white)

$P = 1 - \frac{C_3^2}{C_8^2} = 1 - \frac{3}{28} = \frac{25}{28}$

Answer: $\frac{25}{28}$

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Example 3: Advanced (Difficulty ★★★★☆)

Two people independently solve the same problem. Person A solves with probability 0.7, Person B with 0.8. Find:

1. Probability both solve it
2. Probability exactly one solves it
3. Probability at least one solves it

Solution:

Let $A$ = "A solves", $B$ = "B solves"

Given: $P(A) = 0.7$, $P(B) = 0.8$

(1) Both solve: $P(A \cap B) = P(A) \cdot P(B) = 0.7 \times 0.8 = 0.56$

(2) Exactly one solves: $P = P(A) \cdot P(\overline{B}) + P(\overline{A}) \cdot P(B)$ $= 0.7 \times 0.2 + 0.3 \times 0.8 = 0.14 + 0.24 = 0.38$

(3) At least one solves: Use complement $P = 1 - P(\overline{A}) \cdot P(\overline{B}) = 1 - 0.3 \times 0.2 = 0.94$

Answers: (1) 0.56 (2) 0.38 (3) 0.94

Common Mistakes

❌ Mistake 1: Probability > 1

Correction: Probability range is $[0, 1]$. If result exceeds this, calculation is wrong.

❌ Mistake 2: Using direct method for "at least" problems

Wrong: Enumerate all cases (easy to miss some)

Correct: Use complement method: "at least one" = 1 - "none"

❌ Mistake 3: Confusing independence with mutual exclusivity

Mutually exclusive: Can't occur together ($A \cap B = \emptyset$), $P(A \cup B) = P(A) + P(B)$

Independent: One doesn't affect the other ($P(A \cap B) = P(A) \cdot P(B)$)

Completely different!

❌ Mistake 4: Forgetting conditions in conditional probability

Correction: $P(A|B) \neq P(A)$ (unless $A$ and $B$ are independent)

Study Tips

1. ✅ Master basic concepts: Sample space, elementary events, random events
2. ✅ Memorize formulas: Complement, addition rule, conditional probability
3. ✅ Choose right method: Enumeration for simple, combinatorics for complex
4. ✅ Use complements wisely: "At least" problems are easier with complements
5. ✅ Distinguish independence vs exclusivity: Different definitions and formulas

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💡 Exam Tip: Probability is core CSCA statistics content, accounting for about 40% of statistics problems. Complement method and independent events are frequently tested!