Probability Notation Guide
Before working with formulas, you need to read the notation correctly. Every symbol here has a precise meaning, and confusing them — particularly the union (∪) and intersection (∩), or P(A|B) with P(B|A) — leads to using the wrong formula entirely.
| Symbol | Name | Meaning | Example |
|---|---|---|---|
| P(A) | Probability of A | The likelihood that event A occurs | P(roll a 3) = 1/6 |
| A' or Aᶜ | Complement of A | All outcomes where A does NOT occur | A' = "not rolling a 3" |
| A ∪ B | Union — "A or B" | All outcomes in A, in B, or in both | A ∪ B = "heart or king" |
| A ∩ B | Intersection — "A and B" | Outcomes that belong to both A and B | A ∩ B = "king of hearts" |
| P(A|B) | Conditional probability | Probability of A given B has occurred | P(king | face card) |
| S or Ω | Sample space | The complete set of all possible outcomes | S = {1,2,3,4,5,6} for a die |
| ∅ | Empty set | The impossible event — no outcomes | P(∅) = 0 |
| Σ | Summation | Sum over all specified values | Used in expected value formula |
| n! | Factorial | n × (n−1) × (n−2) × … × 1 | 5! = 120 |
| P(n,r) or nPr | Permutation | Ordered selections of r from n | 10P3 = 720 |
| C(n,r) or nCr | Combination | Unordered selections of r from n | 10C3 = 120 |
Probability Formulas at a Glance
This table summarizes every major probability formula on this page. Use it as your primary reference during problem-solving: identify the problem type in the first column, check the condition is met, then apply the formula.
| Formula Name | Formula | Condition Required | Use When |
|---|---|---|---|
| Basic Probability | P(A) = favorable / total | Equally likely outcomes | Counting favorable outcomes in a finite, equally likely sample space |
| Range Rule | 0 ≤ P(A) ≤ 1 | Always | Validating every probability result — if outside this range, an error was made |
| Complement Rule | P(A') = 1 − P(A) | Always | Finding "not A"; especially for "at least one" problems |
| General Addition | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) | Overlapping events | A or B, when both can occur at the same time |
| Mutually Exclusive Addition | P(A ∪ B) = P(A) + P(B) | P(A ∩ B) = 0 | A or B, when events cannot both occur |
| General Multiplication | P(A ∩ B) = P(A) × P(B|A) | Any events | Both A and B occur — works for dependent or independent events |
| Independent Multiplication | P(A ∩ B) = P(A) × P(B) | Independent events | Both A and B occur, and A's outcome does not affect P(B) |
| Conditional Probability | P(A|B) = P(A ∩ B) / P(B) | P(B) > 0 | Probability of A given B is known to have occurred |
| Bayes' Theorem | P(A|B) = P(B|A) · P(A) / P(B) | P(B) > 0 | Reversing a conditional probability from P(B|A) to P(A|B) |
| Law of Total Probability | P(B) = ΣP(B|Aᵢ)P(Aᵢ) | A₁…Aₙ partition S | Finding P(B) when B's probability depends on which of several cases holds |
| At Least One (general) | 1 − P(none occur) | Any structure | Complement shortcut for "at least one success" |
| At Least One (n indep. trials) | 1 − (1−p)ⁿ | Independent, equal p | At least one success in n independent trials with same success probability p |
| Permutations | P(n,r) = n! / (n−r)! | Order matters | Counting ordered selections of r items from n distinct items |
| Combinations | C(n,r) = n! / [r!(n−r)!] | Order does not matter | Counting unordered selections of r items from n distinct items |
| Expected Value | E(X) = Σ x · P(X = x) | Discrete random variable | Probability-weighted average outcome of a discrete random variable |
| Binomial Probability | P(X=k) = C(n,k) pᵏ(1−p)ⁿ⁻ᵏ | n indep. trials, const. p | Exactly k successes in n independent binary trials |
OR problems use addition formulas. AND problems use multiplication formulas. The follow-up question — which kind of OR or AND? — is what determines the exact version of the rule to apply. Never use the simplified version (no subtraction / no conditional) before confirming the required condition holds.
Basic Probability Formula
A = event of interestS = sample space (all possible outcomes)|A| = number of outcomes in AThis is the formula most students encounter first, but it carries a significant restriction. The direct counting form — favorable divided by total — is valid only when every outcome in the sample space has the same probability. A fair die, a shuffled deck, or a bag of identically-sized marbles all satisfy this condition. A weighted die, or drawing from a bag where each marble is a different size, does not.
P(A) = favorable/total is a counting formula for uniform sample spaces. When the outcomes have different probabilities — a biased coin, for example — you cannot count and divide. You must use the probabilities directly. This is one of the most frequent beginner errors in probability.
The probability of the entire sample space is always 1: P(S) = 1. The probability of the impossible event is always 0: P(∅) = 0. Every probability sits between these extremes: 0 ≤ P(A) ≤ 1 for any event A.
Problem: What is the probability of rolling an even number on a fair six-sided die?
Identify the sample space. S = {1, 2, 3, 4, 5, 6}. Total outcomes: 6. Each outcome is equally likely on a fair die.
Identify the favorable outcomes. Even numbers in S: {2, 4, 6}. Favorable outcomes: 3.
Apply the formula. P(even) = 3 / 6 = 1/2 = 0.5
✓ P(rolling an even number) = 1/2 = 0.5 (50%). Three of the six equally likely outcomes are even.
Complement Rule Formula
A' = complement of A = "not A"P(A) + P(A') = 1 alwaysSince every outcome either belongs to event A or does not belong to event A — with no middle ground and nothing left out — the probabilities of A and its complement must add to exactly 1. This makes the complement rule one of the most efficient shortcuts in probability. It is particularly powerful for "at least one" problems, where computing the complement (zero successes) is often far simpler than listing every way at least one success can occur.
Problem: A quality-control check finds that 15% of parts are defective. What is the probability a randomly selected part passes?
Identify P(A). P(defective) = 0.15
Apply the complement. P(not defective) = 1 − 0.15 = 0.85
✓ P(part passes) = 0.85 (85%). P(defective) + P(not defective) = 0.15 + 0.85 = 1.00 ✓
Formula for "At Least One"
Problem: A fair coin is flipped four independent times. What is the probability of getting at least one head?
Define the complement. P(at least one head) = 1 − P(zero heads) = 1 − P(all tails).
Calculate P(all tails). Tosses are independent, each P(tail) = 1/2. P(all tails) = (1/2)⁴ = 1/16.
Apply the complement. P(at least one head) = 1 − 1/16 = 15/16.
✓ P(at least one head in 4 flips) = 15/16 ≈ 0.9375 (93.75%). This is far simpler than listing all 15 favorable outcomes individually.
Addition Rule Formulas
General Addition Rule
∪ = union = "OR"∩ = intersection = outcomes in both A and BWhen two events share possible outcomes, simply adding P(A) + P(B) counts every outcome that belongs to both events twice — once for A and once for B. The general addition rule corrects this by subtracting P(A ∩ B) once. This is the formula to reach for whenever the problem uses the word "or" and the events are not mutually exclusive.
Using P(A) + P(B) without subtracting P(A ∩ B) when the events overlap. The result exceeds the true probability — and can even exceed 1, which immediately signals an error. Always ask: can A and B occur on the same trial? If yes, you must subtract the intersection.
Problem: What is the probability of drawing a heart or a king from a standard 52-card deck?
Identify probabilities. P(heart) = 13/52. P(king) = 4/52.
Find the intersection. A card can be both a heart and a king — the king of hearts. P(heart ∩ king) = 1/52.
Apply the general addition rule. P(heart ∪ king) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.
✓ P(heart or king) = 16/52 = 4/13 ≈ 0.308 (30.8%). Direct count: 13 hearts + 3 non-heart kings = 16 favorable cards ✓
Mutually Exclusive Events Formula
Problem: Rolling a 2 or a 5 on a single fair die.
Check mutual exclusivity. A single die roll shows exactly one face. You cannot get both 2 and 5 simultaneously, so P(2 ∩ 5) = 0.
Apply the mutually exclusive formula. P(2 or 5) = P(2) + P(5) = 1/6 + 1/6 = 2/6 = 1/3.
✓ P(rolling a 2 or 5) = 2/6 = 1/3 ≈ 0.333 (33.3%). Two of six equally likely outcomes are favorable ✓
Multiplication Rule Formulas
General Multiplication Rule (Dependent Events)
P(B|A) = probability of B given A has occurredThe general multiplication rule works for any two events — dependent or independent. When the outcome of the first event changes the probability of the second, P(B|A) reflects that updated probability. The most common scenario: drawing items from a set without replacing them between draws, so the pool shrinks and the probabilities shift.
Problem: Drawing two queens consecutively from a standard 52-card deck without replacement.
First draw. 4 queens in 52 cards. P(queen on 1st draw) = 4/52 = 1/13.
Update for the second draw. After removing one queen, the deck has 51 cards and 3 remaining queens. P(queen on 2nd | queen on 1st) = 3/51 = 1/17.
Apply the general multiplication rule. P(queen, then queen) = 4/52 × 3/51 = 12/2652 = 1/221.
✓ P(two consecutive queens, no replacement) = 1/221 ≈ 0.00452 (0.452%). Using 4/52 × 4/52 = 16/2704 ≈ 0.59% would be wrong — that ignores the change in deck size.
Independent Events Formula
Problem: Flipping heads AND rolling a 6 at the same time.
Confirm independence. A coin and a die are separate experiments. The coin result gives no information about the die result — they are independent.
Find individual probabilities. P(heads) = 1/2. P(6) = 1/6.
Multiply. P(heads AND 6) = 1/2 × 1/6 = 1/12.
✓ P(heads and rolling 6) = 1/12 ≈ 0.083 (8.3%). The sample space has 12 equally likely outcomes (2 coin × 6 die), and exactly 1 is (H, 6) ✓
Conditional Probability Formula
P(A|B) = "probability of A given B"P(A|B) ≠ P(B|A) in generalSwapping the conditioning event — treating P(A|B) as equal to P(B|A) — is called the base-rate fallacy or the prosecutor's fallacy, and it appears in real legal and medical reasoning errors. Bayes' theorem provides the mathematically correct way to convert between the two.
Problem: From a standard deck, given that the card drawn is a face card, what is the probability it is a king?
Identify the new sample space. Given the card is a face card, we restrict to the 12 face cards (4 jacks, 4 queens, 4 kings).
Count favorable outcomes in the restricted space. 4 kings among 12 face cards. P(king ∩ face card) = 4/52. P(face card) = 12/52.
Apply the conditional formula. P(king | face card) = (4/52) / (12/52) = 4/12 = 1/3.
✓ P(king | face card) = 1/3 ≈ 0.333 (33.3%). Four of the 12 face cards are kings ✓
Bayes' Theorem Formula
P(A) = prior probability of AP(B|A) = likelihood of evidence B given AP(A|B) = posterior probability of A given evidence BBayes' theorem is not a separate branch of probability — it follows directly from the definition of conditional probability. When A and its complement A' partition the sample space, the denominator P(B) can be expanded using the law of total probability:
Problem: A medical test is 95% accurate for a disease that affects 1% of the population. A patient tests positive. What is the probability they actually have the disease?
Define events and identify probabilities. D = has disease. + = tests positive. P(D) = 0.01. P(+|D) = 0.95 (true positive rate). P(+|D') = 0.05 (false positive rate, assuming 95% accurate means 5% false positive).
Calculate P(+) using total probability. P(+) = P(+|D)·P(D) + P(+|D')·P(D') = 0.95×0.01 + 0.05×0.99 = 0.0095 + 0.0495 = 0.059.
Apply Bayes' theorem. P(D|+) = P(+|D)·P(D) / P(+) = 0.0095 / 0.059 ≈ 0.161.
✓ P(disease | positive test) ≈ 16.1%. Despite the test being 95% accurate, the low prevalence (1%) means most positive results are still false positives. This counterintuitive result is why Bayes' theorem matters in medical diagnosis.
For a full treatment of this formula and its applications, see the dedicated Bayes' theorem page, or use the Bayes' Theorem Calculator to verify your own calculations.
Counting Formulas: Permutations and Combinations
Permutations and combinations are counting formulas, not probability formulas by themselves. They become probability formulas when you use them to count the favorable outcomes and the total outcomes inside P(A) = |A| / |S|.
Permutation Formula
n = total number of distinct itemsr = number being selectedn! = n × (n−1) × … × 10! = 1 by definitionProblem: How many ways can a president, vice-president, and secretary be chosen from a group of 10 people?
Identify the structure. The same three people in a different order produces a different outcome (person A as president vs. secretary are different roles). Order matters — use permutations.
Apply the formula. P(10, 3) = 10! / (10−3)! = 10! / 7! = 10 × 9 × 8 = 720.
✓ There are 720 distinct ways to fill the three roles from 10 candidates. Use the Permutation Calculator to verify.
Combination Formula
The difference between permutations and combinations comes down to one question: does the order of selection create a meaningfully different outcome? Choosing president, vice-president, and secretary creates different outcomes depending on who is assigned which role — permutations. Choosing a three-person committee produces the same group regardless of the order people were selected — combinations.
C(10, 3) = 10! / [3! × 7!] = 720 / 6 = 120. The same 10 people yield 720 ordered arrangements but only 120 distinct groups. For more practice, see permutations and combinations or the Combination Calculator.
Expected Value Formula
X = the random variablex = each possible value X can takeP(X = x) = probability that X takes value xThe expected value represents the long-run average outcome of a random variable, not a prediction about any single trial. It need not itself be an observable outcome — the expected value of a fair die roll is 3.5, which no single roll can produce. For the linearity property: E(aX + b) = a·E(X) + b, and E(X + Y) = E(X) + E(Y) without requiring independence. Visit the dedicated expected value page, or use the Expected Value Calculator.
Binomial Probability Formula
n = number of trialsk = number of successesp = probability of success per trialE(X) = np Var(X) = np(1−p)All four conditions must hold before this formula applies: fixed number of trials, each trial is independent, the probability of success is the same on every trial, and each trial has exactly two outcomes (success or failure). Violating any of these requires a different distribution. See the full binomial distribution page for examples, tables, and the calculator.
Which Probability Formula Should You Use?
Use this decision tree as a systematic guide. The path from your problem's structure to the correct formula takes two steps: identify the key word or structure, then confirm the required condition holds before applying the formula.
🔀 Probability Formula Selector
Can A and B happen at the same time?
Does the first event change the probability of the second?
For "at least one": 1 − P(none). Simplified to 1−(1−p)ⁿ only for n independent equal-p trials.
To reverse the conditional — convert P(B|A) to P(A|B) — use Bayes' theorem.
Order doesn't matter → Combinations
💡 Language cues help identify the formula, but always confirm the required condition. "Without replacement" → almost always dependent. "With replacement" → almost always independent.
All Probability Formulas — Quick Reference Cards
Basic Probability
P(A) = |A| / |S|
Favorable outcomes divided by total outcomes. Valid only when all outcomes are equally likely.
Condition: Equally likely outcomesComplement Rule
P(A') = 1 − P(A)
Probability of "not A". P(A) + P(A') = 1 always. Powerful shortcut for "at least one" problems.
Condition: AlwaysGeneral Addition Rule
P(A∪B)=P(A)+P(B)−P(A∩B)
A or B when events can both occur. Subtract overlap to prevent double-counting.
Condition: Overlapping eventsMutually Exclusive Addition
P(A∪B) = P(A) + P(B)
A or B when events cannot both occur. The intersection term is zero, so it disappears.
Condition: P(A∩B) = 0General Multiplication
P(A∩B) = P(A) × P(B|A)
Both A and B occur. Works for any events — dependent or independent. Always safe to use.
Condition: Always (general form)Independent Multiplication
P(A∩B) = P(A) × P(B)
Both A and B occur, and A's outcome does not affect P(B). Simplification of Formula 5.
Condition: Independent eventsConditional Probability
P(A|B) = P(A∩B) / P(B)
Probability of A given B occurred. Restricts the sample space to B.
Condition: P(B) > 0Bayes' Theorem
P(A|B) = P(B|A)·P(A) / P(B)
Reverses a conditional. Converts P(B|A) into P(A|B) using prior probabilities.
Condition: P(B) > 0Permutations
P(n,r) = n! / (n−r)!
Ordered selections. Use when different arrangements count as different outcomes.
Condition: Order mattersCombinations
C(n,r) = n! / [r!(n−r)!]
Unordered selections. Use when the same group in a different order is the same outcome.
Condition: Order doesn't matterExpected Value
E(X) = Σ x · P(X = x)
Probability-weighted average outcome of a discrete random variable.
Condition: Discrete distributionBinomial Probability
P(X=k) = C(n,k)·pᵏ·(1−p)ⁿ⁻ᵏ
Exactly k successes in n independent binary trials with constant success probability p.
Condition: 4 binomial requirementsProbability Formula Calculator
Select a formula below, enter your values, and get the result with a full step-by-step breakdown. For more specialized calculations, the site's Probability Calculator, Conditional Probability Calculator, and Bayes' Theorem Calculator handle more complex inputs.
Probability Formulas — Quick Calculator
Enter P(A) to find P(A') = 1 − P(A)
General Addition Rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). Enter 0 for P(A ∩ B) if events are mutually exclusive.
For independent events enter P(B). For dependent events enter P(B|A) — the updated probability after A occurred.
C(n, r) = n! / [r!(n−r)!] — number of unordered selections of r items from n distinct items.
Common Probability Formula Mistakes
These are the errors that consistently cost marks. Each one traces back to applying a formula without checking whether its condition holds.
| Mistake | Wrong Application | Correct Application | Formula Violated |
|---|---|---|---|
| Using counting formula on unequal outcomes | P(A) = favorable/total for a biased coin | Use the given probabilities directly | Basic probability condition |
| Forgetting to subtract intersection (OR) | P(heart or king) = 13/52 + 4/52 = 17/52 | P = 13/52 + 4/52 − 1/52 = 16/52 | General addition rule |
| Using independent formula on dependent draws | P(two queens) = 4/52 × 4/52 | P = 4/52 × 3/51 = 1/221 | General multiplication rule |
| Swapping P(A|B) and P(B|A) | P(disease|positive) ≈ P(positive|disease) = 95% | Use Bayes' theorem — result ≈ 16.1% (see Example 9) | Bayes' theorem |
| Using (1−p)ⁿ on dependent trials | P(at least one ace in 3 draws, no replacement) = 1−(48/52)³ | Calculate P(no aces) using conditional structure: 48/52 × 47/51 × 46/50 | At least one formula conditions |
| Using permutations when order doesn't matter | Committee of 3 from 10 = P(10,3) = 720 | C(10,3) = 120 (same group, different order = same committee) | Combinations vs. permutations |
| Treating mutually exclusive as independent | Rolling 3 and rolling 5 are independent, so P(3 ∩ 5) = P(3)×P(5) | They are mutually exclusive (P(3 ∩ 5) = 0), therefore dependent, not independent | Independence definition |
| Probability exceeding 1 | P(A ∪ B) = 0.8 + 0.6 = 1.4 | P(A ∪ B) = 0.8 + 0.6 − P(A ∩ B) ≤ 1.0 | Range rule |
Independence vs. Mutual Exclusivity — A Critical Distinction
Mutually exclusive events cannot both occur: P(A ∩ B) = 0. Their Venn diagram circles do not overlap. If you know one occurred, the other definitely did not — so knowing one tells you something about the other.
Independent events can both occur, but one's outcome carries no information about the other: P(B|A) = P(B). Their Venn diagram circles can overlap freely.
Two events with positive probability that are mutually exclusive are, by this reasoning, actually dependent — learning that A occurred changes P(B) from some positive value to zero. This is the exact opposite of independence.
The formulas reflect this distinction directly. Mutual exclusivity simplifies the addition rule. Independence simplifies the multiplication rule. Applying the wrong simplification — using P(A) + P(B) for overlapping events, or P(A)×P(B) for dependent events — is what produces wrong answers. For more, see mutually exclusive events and the probability rules page.
Printable Probability Formulas Cheat Sheet
This condensed reference contains every formula with its condition in a format designed for quick lookup during problem-solving or exam review.
- Basic: P(A) = |A| / |S| — equally likely outcomes only
- Range: 0 ≤ P(A) ≤ 1 — all probabilities, always
- Certain / Impossible: P(S) = 1 | P(∅) = 0
- Complement: P(A') = 1 − P(A)
- At least one (general): 1 − P(none)
- At least one (n indep. trials): 1 − (1−p)ⁿ
- General addition (overlapping): P(A∪B) = P(A)+P(B)−P(A∩B)
- Addition (mutually exclusive): P(A∪B) = P(A)+P(B)
- General multiplication: P(A∩B) = P(A) × P(B|A)
- Multiplication (independent): P(A∩B) = P(A) × P(B)
- Conditional probability: P(A|B) = P(A∩B) / P(B)
- Bayes' theorem: P(A|B) = P(B|A)·P(A) / P(B)
- Law of total probability: P(B) = ΣP(B|Aᵢ)·P(Aᵢ)
- Permutations (order matters): P(n,r) = n! / (n−r)!
- Combinations (order doesn't matter): C(n,r) = n! / [r!(n−r)!]
- Expected value: E(X) = Σ x·P(X=x)
- Binomial: P(X=k) = C(n,k)·pᵏ·(1−p)ⁿ⁻ᵏ
- Binomial mean / variance: E(X) = np | Var(X) = np(1−p)
- OR → Addition | AND → Multiplication | GIVEN → Conditional
Where to Go From Here
Each formula on this page connects to a deeper resource on this site. Use the links below to move from formula recognition to full conceptual mastery.
Frequently Asked Questions
What is the basic probability formula?
P(A) = number of favorable outcomes / total number of possible outcomes. This counting formula applies only when all outcomes in the sample space are equally likely — a fair die, a shuffled standard deck, randomly selected items from a uniform group. If the outcomes have different probabilities, you cannot simply count and divide.
When do I add probabilities vs. multiply them?
Add when the problem asks for A OR B. Multiply when it asks for A AND B. Then confirm the condition: for OR, check whether events overlap (if yes, subtract the intersection). For AND, check whether events are independent (if not, use the conditional form P(A) × P(B|A)).
What is the difference between mutually exclusive and independent events?
Mutually exclusive: P(A ∩ B) = 0 — they cannot both occur. Independent: P(B|A) = P(B) — knowing A's outcome tells you nothing about B. Two positive-probability mutually exclusive events are dependent, because learning one occurred changes the other's probability from positive to zero.
Can probability ever be greater than 1 or less than 0?
No. By the Range Rule, every probability must satisfy 0 ≤ P(A) ≤ 1. A result outside this range signals a calculation error — most often, forgetting to subtract P(A ∩ B) in the general addition rule, causing the result to exceed 1.
Why does the "at least one" formula use 1 minus instead of adding?
Because the complement of "at least one success" is "zero successes," which is usually a single calculation. Without the complement, you would need to list every combination of one success, two successes, three successes, and so on — far more work. The simplified form 1 − (1−p)ⁿ requires independent trials with the same probability p on each.
What is the formula for dependent events?
P(A ∩ B) = P(A) × P(B|A). The key is that P(B|A) — the probability of B given that A has already occurred — reflects the updated conditions after the first event. In drawing without replacement, this means using a smaller pool and fewer items of the target type for the second draw.
How do permutations relate to probability?
Permutations count the number of ordered arrangements, not probabilities by themselves. They become probability tools when you use them inside P(A) = |A| / |S| — for example, counting how many favorable ordered arrangements exist and dividing by the total number of ordered arrangements in the sample space. The same applies to combinations for unordered selections.