General Probability

You are given probabilities for two or three events and asked for the probability of a union or a region described by overlapping conditions.

Start by drawing a Venn diagram so you can label every disjoint region. Apply inclusion-exclusion for unions, and translate the verbal description into a region using complements, intersections, and the universal set. Re-read the question to confirm whether "or" is inclusive and whether the problem asks for "exactly one" versus "at least one".

P(A or B) = P(A) + P(B) - P(A and B) generalizes by inclusion-exclusion.

A test or population is split into mutually exclusive groups with different prior probabilities and conditional success rates, and you are asked for the probability of belonging to a group given an observed outcome.

Identify the partition first, then label each P(group) and each P(outcome | group). Use total probability to compute the denominator P(outcome), then plug into Bayes for the posterior. Sanity check by confirming the posteriors over all groups sum to one.

P(A | B) = P(B | A) P(A) / sum over partition of P(B | A_i) P(A_i).

You are told two events have certain probabilities and an additional relationship, and you must decide if they are independent or compute a joint probability.

Test independence using P(A and B) = P(A) P(B); equivalently P(A | B) = P(A). Be careful not to confuse independence with mutual exclusivity — mutually exclusive events with positive probability are dependent. Translate the words "given" and "and" into the right symbol before computing.

Independence: P(A and B) = P(A) P(B); mutually exclusive ≠ independent.

A sampling problem asks for the probability of a specific arrangement or selection — e.g., committees, license plates, or hands of cards.

Decide first whether order matters (permutations) or not (combinations), and whether sampling is with or without replacement. Build the numerator as the count of favorable outcomes and the denominator as total outcomes under the same convention. When events have multiple stages, multiply counts; when alternatives are exclusive, add.

C(n, k) = n! / (k! (n - k)!); P(n, k) = n! / (n - k)!.