Skill: probability

Probability Rules & Counting

Apply the addition and multiplication rules, tell independent from dependent events, and count with permutations and combinations.

1 Probability from counts

A probability is a fraction: favorable outcomes over total equally-likely outcomes. It is always between 0 and 1.

Basic rule

P(event) = favorable / total

Read “P(A)” as “the probability of A”. A probability of 0 means impossible; 1 means certain.

Example Marble bag

A bag holds 5 red and 7 blue marbles (12 total).
P(red) = 5/12 ≈ 0.417.

2 The addition rule: P(A or B)

“Or” means at least one happens. If A and B can both happen, the overlap gets counted twice — so subtract it once.

Addition rule

P(A or B) = P(A) + P(B) − P(A and B)

If A and B are mutually exclusive (cannot both happen), the last term is 0 and you just add.

Example Honor roll or athlete

In a class, P(honor roll) = 0.3, P(athlete) = 0.4, P(both) = 0.1.
P(honor roll or athlete) = 0.3 + 0.4 − 0.1 = 0.6.
Forgetting the overlap would give 0.7 — double-counting the 10% who are both.

3 The multiplication rule: P(A and B)

“And” means both happen. For independent events (one does not affect the other), just multiply. Otherwise, adjust for what already happened.

Multiplication rule

Independent: P(A and B) = P(A) × P(B)

Dependent: P(A and B) = P(A) × P(B | A)

P(B | A) reads “probability of B given A happened” — recompute with the new totals. This is called conditional probability.

Example Two coin flips (independent)

P(heads) = 1/2 each flip; flips do not affect each other.
P(heads and heads) = 1/2 × 1/2 = 1/4.

Example Two draws without replacement (dependent)

Bag: 5 red, 7 blue (12 total). Draw two marbles without putting the first back.
P(both red) = 5/12 × 4/11 = 20/132 = 5/33.
After one red is gone, only 4 red of 11 remain — that is P(second red | first red).

4 Counting: permutations vs combinations

When every outcome is equally likely, P = (good arrangements) / (all arrangements). Count arrangements with permutations (order matters) or combinations (order does not).

Counting formulas

P(n, k) = n! / (n−k)! | C(n, k) = n! / (k!(n−k)!)

Order matters → permutation (president, then vice-president). Order does not → combination (a committee of 3).

Example Awards vs committee

Gold/silver/bronze from 8 runners: order matters → P(8,3) = 8×7×6 = 336.
Choose any 3 of 8 for a team: order does not matter → C(8,3) = 336/6 = 56.

5 The complement shortcut

“At least one” problems are often easier backwards: compute the chance of none, then subtract from 1.

Complement rule

P(at least one) = 1 − P(none)

Example At least one six in 3 rolls

P(no six in one roll) = 5/6; three rolls: (5/6)³ = 125/216.
P(at least one six) = 1 − 125/216 = 91/216 ≈ 0.421.

6 Common mistakes

Adding without subtracting the overlap. P(A or B) = P(A) + P(B) only when A and B are mutually exclusive. Otherwise subtract P(A and B) — the Venn diagram never lies.

Permutation when order does not matter. Choosing a committee is a combination; using P(n,k) overcounts by k! (every ordering of the same committee).

Treating dependent events as independent. Drawing without replacement changes the totals — recompute the second probability given the first draw.

7 Key vocabulary

Mutually exclusive: Cannot both happen; P(A and B) = 0.
Independent: One event does not change the other’s probability; multiply.
Conditional probability: P(B | A): probability of B given A happened.
Permutation: Ordered arrangement: P(n,k) = n!/(n−k)!.
Combination: Unordered selection: C(n,k) = n!/(k!(n−k)!).
Complement: P(not A) = 1 − P(A).

8 Quick checks

1. P(A) = 0.5, P(B) = 0.4, P(both) = 0.2. Find P(A or B).

0.7. 0.5 + 0.4 − 0.2 — subtract the overlap.

2. Deal 2 cards from a deck. Why is P(both aces) not (4/52)²?

Dependent draws. After one ace, 3 aces of 51 remain: (4/52)(3/51).

3. Top 3 finishers vs a 3-person committee from 10: which is P(10,3)?

Top 3 finishers. Places are ordered — the committee is C(10,3).

Practice Counting calculator Coin-flip lab