Mathfolis

Probability Rules

Unit 4 · Probability, Random Variables & Distributions

What AP Stats asks here

Three rules carry most of Unit 4 in the first pass: complement, addition with overlap, and Law of Large Numbers. The recurring AP traps are double-counting in the addition rule (forgetting to subtract the overlap) and the gambler's fallacy (assuming streaks shift independent trial probabilities).

Core rules

Range
0≤P(A)≤10 \le P(A) \le 1
Complement
P(Ac)=1−P(A)P(A^c) = 1 - P(A)
Addition with overlap
P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
Disjoint shortcut
P(A∪B)=P(A)+P(B)    when P(A∩B)=0P(A \cup B) = P(A) + P(B) \;\;\text{when } P(A \cap B) = 0

Useful identity

At least one
P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none})
AP Tip: When the question says 'at least one', reach for the complement. Computing P(none) is one shot; summing the cases (exactly 1, exactly 2, ...) is many shots and prone to overlap errors.
Caution: The Law of Large Numbers describes long-run frequencies, not memory. After ten heads, the next fair-coin flip is still 0.5 for tails — past results do not 'owe' future trials.
Type 1

Addition rule and complement

Subtract the overlap once to undo the double-count in P(A)+P(B)P(A) + P(B). For disjoint events the overlap is zero.

Example 1
In a survey, 60% like coffee, 50% like tea, and 30% like both. What is the probability a random respondent likes at least one of the two? (A) 0.50 (B) 0.80 (C) 1.10 (D) 0.30

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Type 2

Law of Large Numbers vs gambler's fallacy

Independent trials have no memory. LLN says the long-run frequency stabilizes, not that the next outcome compensates for a streak.

Example 2
A fair coin has just landed heads on 10 consecutive flips. What is P(tails on the 11th flip)P(\text{tails on the 11th flip})? (A) Greater than 0.5 because tails is overdue. (B) Exactly 0.5 — the coin has no memory and trials are independent. (C) Less than 0.5 because the streak shows heads is more likely. (D) Cannot be determined without more flips.

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Type 3

'At least one' via complement

P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none}) is the cleanest path when the trials are independent.

Example 3
A box has 4 red and 6 blue balls. Two are drawn with replacement. What is P(at least one red)P(\text{at least one red})? (A) 0.36 (B) 0.40 (C) 0.64 (D) 0.80

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