Statistics Quest

Probability rules

The complement, addition and multiplication rules combine probabilities of events; all follow from probabilities being between 0 and 1 and summing to 1 over the sample space.

P(Ac)=1−P(A)P(A^c) = 1 - P(A)
Complement rule.
AcA^c
The event 'not A'.
P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
General addition rule (for mutually exclusive events the last term is 0).
A∪BA \cup B
A or B (or both).
A∩BA \cap B
A and B.
P(A∩B)=P(A) P(B∣A)P(A \cap B) = P(A)\,P(B \given A)
General multiplication rule; equals P(A)P(B) when A and B are independent.
P(B∣A)P(B \given A)
Probability of B given that A occurred.
P(B)=∑kP(B∣Ak) P(Ak)P(B) = \sum_{k} P(B \given A_k)\,P(A_k)
Law of total probability (the A_k partition the sample space).
A1,A2,…A_1, A_2, \ldots
Mutually exclusive events that together cover every outcome.

When to use

  • Combining probabilities of simple events into probabilities of compound events.
  • Probability trees: multiply along branches, add across branches leading to the same outcome.
  • "At least one" questions: P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none}).

Assumptions

Equally likely outcomes only when stated (fair dice, random draws); otherwise probabilities come from a model or long-run relative frequencies.

How to interpret

Every probability is between 0 and 1. The probabilities of all outcomes in the sample space sum to 1. Mutually exclusive events cannot happen together; independent events do not influence each other — these are different properties.

Common mistakes

  • Adding probabilities of events that overlap without subtracting the intersection.
  • Multiplying probabilities of dependent events as if independent.
  • Treating "mutually exclusive" as "independent" (two exclusive events with positive probabilities are always dependent).

Example

P(A)=0.5P(A) = 0.5, P(B)=0.4P(B) = 0.4, P(A∩B)=0.2P(A \cap B) = 0.2: P(A∪B)=0.5+0.4−0.2=0.7P(A \cup B) = 0.5 + 0.4 - 0.2 = 0.7 and P(neither)=0.3P(\text{neither}) = 0.3. Since P(A)P(B)=0.2=P(A∩B)P(A)P(B) = 0.2 = P(A \cap B), A and B happen to be independent.