The methods, techniques and warnings — the things that don't fit on a flashcard. Read it, learn it, use it.
Before you do anything else, ask does the order matter?
If yes → Permutation. (Arrangements. Codes. Race positions. Words.)
If no → Combination. (Choosing a team. Picking a committee. Selecting items.)
AND = multiply. If task 1 has p ways and task 2 has q ways, doing both has p × q ways.
OR = add. If task 1 has p ways and task 2 has q ways, doing one or the other has p + q ways.
If certain people must be together:
1. Glue them into a bubble. Treat the bubble as one object.
2. Arrange the bubbles (plus the other items) normally.
3. Multiply by the arrangements inside each bubble.
“Not together” = Total − Together.
If some of the objects you’re arranging are identical, divide n! by the factorial of each group of identicals.
n!k1! · k2! · …
For codes / numbers / passwords with a restriction (can’t start with 0, must be odd, no repeats):
1. Fill the restricted slot first.
2. Then fill the rest.
3. Watch whether digits can repeat or not — that changes the count for each slot.
Triangles from a set of points: nC3. But subtract any sets of 3 points that are collinear (on a straight line) — they don’t form a triangle.
Quadrilaterals between two parallel lines: choose 2 from each line. mC2 × nC2.
Parallelograms: need equal gaps between the chosen points on each line.
nCr = nCn−r
When r is large, flip it — saves a lot of work.
Also: nC0 = nCn = 1.
Pick whichever fits the question:
1. Logic — count the cases directly.
2. Sample space — list every outcome.
3. Tree diagram — for multi-stage events.
4. Permutations — when order matters.
5. Combinations — when order doesn’t matter.
Almost always faster:
P(at least one) = 1 − P(none)
Counting “at least one” directly means adding 1, 2, 3, … cases. Counting “none” is one case — subtract from 1.
With replacement — probabilities stay the same each draw.
Without replacement — the denominator shrinks each draw (and the numerator may change too).
1. Draw a branch for each outcome at each stage.
2. Write the probability on every branch.
3. Multiply along a branch (AND).
4. Add between branches when more than one path gives the result you want (OR).
E(x) = Σ x · P(x)
Multiply each outcome value by its probability and add them up. Often paired with ΣP(x) = 1 in a two-equation problem.
Fair game ⇒ E(x) = 0.
Game in your favour ⇒ E(x) > 0.
Game against you ⇒ E(x) < 0.
Expected count = P(event) × number of trials.
The multiplication rule for intersection is NOT general.
It only holds when A and B are independent.
For general events, you need the Venn diagram or the formula P(A ∩ B) = P(A | B) · P(B).
Either of these tests works (use whichever is easier from the info given):
1. Check whether P(A | B) = P(A).
2. Check whether P(A) · P(B) = P(A ∩ B).
If yes, independent. If no, dependent.
P(A ∩ B)P(B)
Read it as: “what fraction of B is also A?” The condition B becomes the new total.
In a Venn diagram: shade B — that’s your new universe. Then the overlap with A as a fraction of all of B is P(A | B).
Mutually exclusive — can’t happen together. P(A ∩ B) = 0. (Circles don’t overlap.)
Exhaustive — no other options. P(A ∪ B) = 1. (Nothing sits outside the two circles.)
Independent — one has no effect on the other. (See cards 13 & 14.)
1. Only 2 possible outcomes — success or failure.
2. A fixed number of trials n.
3. Trials are independent.
4. The probability of success is constant on each trial.
P(r) = nCr · pr · qn−r
n = number of trials. r = number of successes you want. p = P(success). q = 1 − p = P(failure).
The nCr piece counts every way to arrange r successes among the n trials — Bernoulli takes care of the order for you.
Two pieces, multiplied together:
1. Bernoulli on the first (n − 1) trials for (r − 1) successes.
2. × p for the success on the n-th trial.
P = n−1Cr−1 · pr−1 · qn−r · p