Probability Rules

Why This Matters

Probability rules form the mathematical backbone of everything you'll do in AP Statistics—from calculating the likelihood of sample outcomes to understanding why confidence intervals and hypothesis tests work the way they do. When you encounter sampling distributions, you're applying probability rules to predict how sample statistics behave. When you calculate the probability of a Type I error, you're using these same foundational concepts. The exam tests whether you can choose the right rule for a given scenario, not just whether you've memorized formulas.

The key distinction you're being tested on is recognizing the structure of a probability problem: Are events independent or dependent? Are they mutually exclusive or overlapping? Can you work backward from a result to find a cause? Don't just memorize formulas—know when each rule applies and why it works. That's what separates a 3 from a 5.

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Combining Probabilities: The Addition Rules

When you need the probability that at least one of two events occurs (A or B), you're in addition rule territory. The key question is whether the events can happen simultaneously.

Addition Rule for Mutually Exclusive Events

• Use when events cannot occur together—if one happens, the other is impossible (like rolling a 2 or rolling a 5 on a single die)
• Formula: $P(A \text{ or } B) = P(A) + P(B)$—no overlap to subtract because there is none
• Watch for language like "distinct outcomes," "cannot both occur," or scenarios with non-overlapping categories

Addition Rule for Non-Mutually Exclusive Events

• Use when events can overlap—both could happen at the same time (like drawing a red card or drawing a king)
• Formula: $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$—subtract the intersection to avoid double-counting
• Most common error: forgetting to subtract the overlap, which inflates your probability (sometimes above 1!)

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Joint Probabilities: The Multiplication Rules

When you need the probability that both events occur (A and B), you're using multiplication. The critical distinction is whether knowing one event changes the probability of the other.

Multiplication Rule for Independent Events

• Use when one event doesn't affect the other—like flipping a coin twice or selecting with replacement
• Formula: $P(A \text{ and } B) = P(A) \times P(B)$—probabilities multiply directly
• Independence test: Check if $P(B|A) = P(B)$; if true, events are independent

Multiplication Rule for Dependent Events

• Use when one event changes the probability of another—like drawing cards without replacement
• Formula: $P(A \text{ and } B) = P(A) \times P(B|A)$—the conditional probability accounts for how A affects B
• Common scenario: sampling without replacement, where the sample space shrinks after each selection

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Conditional Probability and Reversing Direction

These rules handle situations where you have partial information or need to update probabilities based on new evidence. Conditional probability is the foundation for inference.

Conditional Probability

• Definition: The probability of A occurring given that B has already occurred—written as $P(A|B)$
• Formula: $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$—restricts the sample space to only outcomes where B occurred
• Key insight: Conditional probability changes the denominator; you're finding what fraction of B's outcomes also include A

Bayes' Theorem

• Use when you need to reverse conditional direction—you know $P(B|A)$ but need $P(A|B)$
• Formula: $P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}$—updates prior probability with new evidence
• Classic applications: diagnostic testing (disease given positive test), quality control, and any "given this result, what caused it?" question

Law of Total Probability

• Use to find $P(B)$ when B can occur through multiple pathways—breaks a complex event into mutually exclusive cases
• Formula: $P(B) = \sum P(B|A_i) \times P(A_i)$—where the $A_i$ partition the sample space
• Often paired with Bayes' Theorem to calculate the denominator $P(B)$

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Simplifying Strategies: Complements and Counting

These tools make complex probability calculations manageable by reframing the problem or systematically counting outcomes.

Complement Rule

• Use when "at least one" or "none" appears—it's often easier to calculate what you don't want
• Formula: $P(A') = 1 - P(A)$, so $P(\text{at least one}) = 1 - P(\text{none})$
• Why it works: Probabilities must sum to 1, so the complement captures everything the event doesn't

Probability of At Least One Event

• Strategic application of the complement rule for repeated independent trials
• Formula: $P(\text{at least one success in } n \text{ trials}) = 1 - P(\text{all failures}) = 1 - (1-p)^n$
• Common trap: Trying to add probabilities directly instead of using the complement approach

Permutations and Combinations in Probability

• Permutations count arrangements where order matters: $_nP_r = \frac{n!}{(n-r)!}$
• Combinations count selections where order doesn't matter: $_nC_r = \frac{n!}{r!(n-r)!}$
• Probability application: Use counting to find the number of favorable outcomes divided by total outcomes in equally likely sample spaces

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Quick Reference Table

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Self-Check Questions

1. You're told $P(A) = 0.4$, $P(B) = 0.3$, and $P(A \text{ and } B) = 0.12$. Are A and B independent? How do you know, and what rule confirms this?

2. A medical test has a 95% detection rate for a disease (sensitivity) and a 3% false positive rate. If 1% of the population has the disease, what's the probability someone with a positive test actually has the disease? Which rules do you need?

3. Compare the addition rule for mutually exclusive events with the general addition rule. Under what condition does the general rule simplify to the mutually exclusive version?

4. You're drawing 3 cards from a standard deck without replacement. Explain why you must use the dependent multiplication rule, and describe how the 10% condition relates to treating this as approximately independent.

5. An FRQ asks: "Find the probability of getting at least one head in 5 coin flips." Write out both the complement approach and explain why directly calculating $P(1) + P(2) + P(3) + P(4) + P(5)$ is less efficient.