条件概率tiáojiàn gàilǜ

Core Concept

Conditional Probability is the probability that event $A$ occurs given that event $B$ has occurred, denoted $P(A|B)$.

Definition

Let $A, B$ be two events with $P(B) > 0$. The conditional probability of $A$ given $B$ is:

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

Understanding:

• Numerator $P(A \cap B)$: Probability of both $A$ and $B$
• Denominator $P(B)$: Probability of $B$
• Meaning: Probability of $A$ in the "new sample space" where $B$ occurred

Properties

1. Non-negativity

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

2. Certain Event

$P(\Omega|B) = 1$

3. Addition Rule

If $A_1, A_2$ are mutually exclusive: $P(A_1 \cup A_2|B) = P(A_1|B) + P(A_2|B)$

Multiplication Rule

From the definition, we get the multiplication rule:

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

Independence

Independent Events

If events $A$ and $B$ are independent: $P(A|B) = P(A)$ $P(B|A) = P(B)$ $P(A \cap B) = P(A) \cdot P(B)$

Meaning: Occurrence of $B$ doesn't affect probability of $A$.

Mutually Exclusive vs Independent

• Mutually exclusive: $A \cap B = \emptyset$, cannot occur together
• Independent: $P(A \cap B) = P(A) \cdot P(B)$, occurrence is independent

Note: Mutually exclusive events are generally not independent (unless one has probability 0)

CSCA Practice Problems

[Example 1] Basic (Difficulty ★★☆☆☆)

A bag has 3 red and 2 white balls. Draw 2 balls without replacement:

1. Probability first is red: $P(A)$
2. Given first is red, probability second is red: $P(B|A)$

Solution:

1. $P(A) = \frac{3}{5}$

2. After drawing red, 2 red and 2 white remain: $P(B|A) = \frac{2}{4} = \frac{1}{2}$

Answer: $P(A) = \frac{3}{5}$, $P(B|A) = \frac{1}{2}$

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[Example 2] Intermediate (Difficulty ★★★☆☆)

Flip two fair coins. Let:

• $A$: At least one heads
• $B$: Both heads

Find $P(B|A)$.

Solution:

Sample space: $\Omega = \{HH, HT, TH, TT\}$

Event $A$: At least one heads, $A = \{HH, HT, TH\}$, $P(A) = \frac{3}{4}$

Event $B$: Both heads, $B = \{HH\}$, $P(B) = \frac{1}{4}$

Intersection: $A \cap B = \{HH\}$, $P(A \cap B) = \frac{1}{4}$

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1/4}{3/4} = \frac{1}{3}$

Answer: $\frac{1}{3}$

Bayes' Theorem

$P(B_i|A) = \frac{P(B_i) \cdot P(A|B_i)}{\sum_{j=1}^{n} P(B_j) \cdot P(A|B_j)}$

Application: Find probability of "cause" given "effect"

Common Misconceptions

❌ Misconception 1: Wrong formula

Wrong: $P(A|B) = \frac{P(A)}{P(B)}$

Correct: $P(A|B) = \frac{P(A \cap B)}{P(B)}$

❌ Misconception 2: Confusing $P(A|B)$ and $P(B|A)$

Wrong: Thinking $P(A|B) = P(B|A)$

Correct: Generally $P(A|B) \neq P(B|A)$

Study Tips

1. ✅ Understand definition: Probability in new sample space
2. ✅ Master formula: $P(A|B) = \frac{P(A \cap B)}{P(B)}$
3. ✅ Distinguish concepts: Mutually exclusive vs independent
4. ✅ Use tree diagrams: Helps visualize complex problems

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💡 Exam Tip: Conditional probability is key in CSCA probability! Must understand deeply. Accounts for 30-40% of probability problems.