Probability puts a number on uncertainty, from the chance of rain to the odds in a card game. A probability is always between 0 (impossible) and 1 (certain). This guide covers the rules you need for most high school and intro college courses, each with a worked example.
The basic formula
When all outcomes are equally likely:
P(event) = number of favorable outcomes / total number of outcomes
Rolling a fair six-sided die, the chance of a number greater than 4 is 2/6 = 1/3, because two outcomes (5 and 6) are favorable.
The complement of an event is everything else: P(not A) = 1 − P(A). This is often the easiest route. The chance of not rolling a 6 is 1 − 1/6 = 5/6.
The addition rule: "or"
P(A or B) = P(A) + P(B) − P(A and B)
You subtract the overlap so outcomes in both events are not counted twice. If the events cannot happen together (they are mutually exclusive), the overlap is 0.
Worked example 1: cards
From a standard 52-card deck, what is the probability of drawing a heart or a king?
- P(heart) = 13/52 and P(king) = 4/52.
- One card is both, the king of hearts: P(both) = 1/52.
- P(heart or king) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
The multiplication rule: "and"
For independent events (one does not affect the other): P(A and B) = P(A) × P(B).
For dependent events: P(A and B) = P(A) × P(B given A), where the second probability accounts for what already happened.
Worked example 2: independent vs. dependent
A bag has 5 red and 3 blue marbles. Find the probability of drawing two reds.
- With replacement (independent): (5/8) × (5/8) = 25/64 ≈ 0.39.
- Without replacement (dependent): (5/8) × (4/7) = 20/56 = 5/14 ≈ 0.36. After one red is removed, 4 reds remain out of 7.
Worked example 3: "at least one"
What is the probability of getting at least one 6 in four rolls of a die?
Use the complement. The chance of no 6 in four rolls is (5/6)⁴ ≈ 0.482, so the chance of at least one 6 is 1 − 0.482 ≈ 0.518.
Counting outcomes
- Fundamental counting principle: if one choice has m options and another has n, together there are m × n. Three shirts and four pants make 12 outfits.
- Permutations (order matters):
nPr = n! / (n − r)!. Ways to award gold, silver and bronze among 8 runners: 8!/5! = 8 × 7 × 6 = 336. - Combinations (order does not matter):
nCr = n! / (r!(n − r)!). Ways to choose a 3-person committee from 8 people: 336 / 3! = 56.
Ask yourself: if I rearranged the selection, would it count as different? If yes, use permutations; if no, use combinations.
Expected value
Expected value is the long-run average of a random outcome: multiply each outcome by its probability and add. A game that pays 10 dollars with probability 0.2 and 0 otherwise, costing 3 dollars to play, has expected value 0.2 × 10 − 3 = −1 dollar per play, so you lose on average.
Common mistakes
- Adding probabilities for "and" events, or multiplying for "or" events.
- Forgetting to subtract the overlap in the addition rule.
- Treating draws without replacement as independent.
- The gambler's fallacy: thinking a coin is "due" for heads after several tails. Each fair flip is still 1/2.
Practice questions
- What is the probability of rolling a sum of 7 with two dice?
- Two cards are drawn without replacement. What is the probability both are aces?
- How many ways can 5 books be arranged on a shelf?
- How many 2-topping pizzas can be made from 6 toppings?
Answers: (1) 6/36 = 1/6. (2) (4/52)(3/51) = 1/221. (3) 5! = 120. (4) 6C2 = 15.
Last reviewed 2026-10-23 by the BrainBoost team.