The Five Probability Rules You Actually Need
| Situation | Formula |
|---|---|
| One event | P(A) = favourable ÷ total outcomes (always between 0 and 1) |
| Not A | P(A′) = 1 − P(A) |
| A or B | P(A or B) = P(A) + P(B) − P(A and B) |
| A and B (independent) | P(A and B) = P(A) × P(B) |
| A and B (dependent) | P(A and B) = P(A) × P(B | A) |
Everything else on this page (at least one, Bayes, conditional chances) is built from these rows.
Independent, Mutually Exclusive or Dependent?
| Type | Meaning | Example | P(A and B) |
|---|---|---|---|
| Independent | One does not change the other’s chance | Two coin tosses | P(A) × P(B) |
| Mutually exclusive | Cannot both happen | Rolling a 2 and a 5 on one die | 0 |
| Dependent | One changes the other’s chance | Two cards drawn without replacement | P(A) × P(B | A) |
Mutually exclusive does not mean independent. If A and B cannot both happen, learning that A occurred tells you B did not, so they are strongly dependent. This is the most frequently confused pair in introductory courses.
Single Events: Dice, Coins and Cards
Count favourable outcomes and divide by all equally likely outcomes. A heart from a 52-card deck is 13 ÷ 52 = 1 ⁄ 4. A six on a fair die is 1 ⁄ 6 or 16.67%. Odds are a different scale: 1 ⁄ 6 is odds of 1 : 5 for, or 5 : 1 against. To count outcomes in harder situations, use the permutation and combination calculator. For example, a 6-from-49 lottery has 13,983,816 equally likely tickets, so one ticket wins the jackpot with probability 1 in 13,983,816.
“At Least One”: The Shortcut That Saves Time
“At least one” is nearly always easiest through the complement: P(at least one) = 1 − P(none). Rolling a die four times, the chance of at least one six is 1 − (5⁄6)⁴ = 51.77%, slightly better than even. That is the classic problem that the Chevalier de Méré asked Pascal about in the 1650s, and it helped start probability theory. The calculator’s “Repeated trials” tab also tells you how many tries you need to reach, say, 90% confidence of at least one success. For a 1⁄6 chance it is 13 rolls.
Conditional Probability and Bayes’ Theorem: A Test Example
Suppose a condition affects 1% of people. A test catches 90% of true cases (sensitivity) but wrongly flags 9% of healthy people (false positives). If you test positive, what is the chance you really have it?
| Of 10,000 people | True cases | Healthy | Total |
|---|---|---|---|
| Test positive | 90 | 891 | 981 |
| Test negative | 10 | 9,009 | 9,019 |
| All | 100 | 9,900 | 10,000 |
Only 90 of the 981 positives are real: 9.2%. Most people guess about 90%. The reason is the base rate: healthy people vastly outnumber sick ones, so even a small false-positive rate produces more false alarms than true hits. Switch to the Bayes tab and try a rare condition with an accurate test. Our view: the single most underrated skill in everyday statistics is thinking in natural frequencies (“out of 10,000 people…”) instead of percentages. It makes this result obvious.
Streaks, Coincidences and the Gambler’s Fallacy
Exact calculations beat intuition here. In 20 fair coin flips the chance of a run of five or more heads somewhere is 25%, and in 100 flips the chance of a run of seven or more is 31.8%. Streaks are normal. Likewise, in a group of 23 people the chance that two share a birthday is 50.7%. None of this means the next flip is “due” to change. Each independent flip is still 50 : 50, and that is the gambler’s fallacy.
Experimental vs Theoretical Probability
Theoretical probability comes from counting equally likely outcomes. Experimental probability is observed frequency, successes ÷ trials. By the law of large numbers the second approaches the first as trials increase, but slowly: the typical error shrinks like 1 ÷ √n, so 100 trials gives about ±5 percentage points near 50%, and you need about 10,000 trials to be within one point.
Common Mistakes
- Adding P(A) + P(B) for events that can overlap, without subtracting P(A and B).
- Multiplying probabilities for dependent events as if they were independent.
- Using “1 − P(A)” when the question asks for “at least one” of several different events.
- Forgetting to change the probabilities when drawing without replacement.
- Confusing P(A | B) with P(B | A), the heart of the prosecutor’s fallacy.
Related: the z-score calculator for normal-distribution probabilities and the standard deviation calculator for the spread.
Why Students Get Stuck on Probability, and What a Good Calculator Should Do
Searching for help with probability usually starts late at night with one problem that will not work out. "and", "or" and "at least one" each need their own rule. A good calculator fixes that by showing each step, checking its own answer and explaining the idea in plain words. This section covers what students struggle with, what they want, and how this page responds.
In short: students need to pick the right rule and keep every outcome in view, not just see a final number, and a calculator earns trust only when its steps can be checked.
Understand the pain points
What students get stuck on
Most mistakes here are not about arithmetic. They come from one rule applied the wrong way round, which is why seeing every step matters.
- Adding probabilities that should be multipliedProbability of rolling a 4 or a 5 on a fair die↗
- Forgetting to subtract the overlap in "or"Probability of drawing 2 aces from a deck without replacement↗
- Mixing up independent and dependent eventsProbability of rolling a sum of 7 with two dice↗
- Struggling with "at least one" questions
Understand the frustrations and desires
What they feel, and what they want
Each frustration points to a wish: a clearer step, a built-in check, an explanation in plain words.
- Wants the rule named and justified
- Wants "at least one" shown with the complement
- Wants conditional probability and Bayes explained
Build the page around their intent
Each intent has a place on this page
The page follows intent, not features, so each need has a clear starting point.
- I need an answer now: type the problem into the calculator at the top of this page.Probability of rolling a 4 or a 5 on a fair die↗
- I need to understand: the sections below explain the idea in plain words and list the common mistakes.
- I need to check my work: every step is shown, so the first line that differs from yours is the one to study.
- I need practice: open the solver, tap + and choose Create Practice Test.
- I need it explained aloud: in the solver, tap + and choose Create Video.
Our View: What Is Overrated, Underrated and Wrong About Probability
Overrated. Memorising a separate formula for every case: union, intersection, conditional. They are the same counting idea written in different shapes.
Underrated. Drawing a tree or a table, then checking that all the branches add up to 1. It catches a missing case faster than any formula, and it works for conditional problems too.
Plain wrong. The gambler’s fallacy: “five heads in a row, so tails is due.” A fair coin has no memory, so the next flip is still 1/2. Another common slip is adding P(A) and P(B) for “A or B” when the events can happen together; that needs P(A) + P(B) − P(A and B).
Test it: for one fair die, P(even or greater than 3) should be 4/6, not 3/6 + 3/6.