Summary: This probability calculator computes events and normal CDF with live steps, formulas and a chart. It accepts labeled numeric inputs, works offline through file:// and includes source-backed explanations for students, analysts and researchers.
Probability calculator for union, intersection, conditional probability, complement and normal CDF with formulas and Venn chart. The calculator works offline, updates instantly and includes a worked example, plain-text formula, MathML, references and structured data.
Default values are loaded. Click any field and edit it; results and chart update automatically.
| Input or setting | Result or interpretation | Use this when |
|---|---|---|
| P(A)=0.4 | complement = 0.6 | event probabilities |
| independent A and B | P(A and B)=P(A)P(B) | event probabilities |
normal 60| about 0.683 | event probabilities | |
P(A union B)=P(A)+P(B)-P(A and B); P(A|B)=P(A and B)/P(B); normal area = Phi(z_b)-Phi(z_a)Probability formulas combine event probabilities, conditional probability and normal CDF areas.
P(A)=0.4, P(B)=0.5, independent
P(A and B)=0.2; P(A union B)=0.7
This page uses the shared statistics core for distribution functions, quantiles and exact integer counting where needed. The formula is shown in plain text so screen readers and search engines can parse it reliably.
Use Load example in the calculator to reproduce this reference result.
{
"tool": "Probability Calculator",
"input": "P(A)=0.4, P(B)=0.5, independent",
"output": "P(A and B)=0.2; P(A union B)=0.7",
"formula": "P(A union B)=P(A)+P(B)-P(A and B); P(A|B)=P(A and B)/P(B); normal area = Phi(z_b)-Phi(z_a)"
}| Calculator | Example input | Expected output |
|---|---|---|
| Probability Calculator | P(A)=0.4, P(B)=0.5, independent | P(A and B)=0.2; P(A union B)=0.7 |
What does p = 0.03 mean? If the null hypothesis and model assumptions were true, a result at least this extreme would occur about 3% of the time. The American Statistical Association cautions that a p-value alone does not measure effect size, practical importance or the probability that Hâ‚€ is true.3
For most classroom and professional reports, pair the calculator result with the question you are answering. A mean or median summarizes location, but spread explains consistency. A confidence interval estimates plausible values, while a hypothesis test evaluates compatibility with a null model. Regression and correlation describe association, so they should be reported with a chart and residual or outlier review. When a result is statistically significant, still ask whether the effect is large enough to matter in the real setting.
| Statistic | Small | Medium | Large | Use |
|---|---|---|---|---|
| Cohen's d | 0.2 | 0.5 | 0.8 | t-test effect size |
| Cramér's V | 0.1 | 0.3 | 0.5 | chi-square association |
| |r| | 0.10 | 0.30 | 0.50 | correlation strength |
| R² | 0.01 | 0.09 | 0.25 | variance explained |
For equally likely outcomes, P(event) = number of favourable outcomes / total possible outcomes. Rolling a fair die and asking for P(rolling a 4): one favourable outcome out of six, so P = 1/6 ≈ 0.167. For drawing a heart from a standard deck: 13 hearts out of 52 cards, P = 13/52 = 0.25. Always check that outcomes are equally likely before using this rule. In real-world settings where outcomes aren't equal, you'd estimate probabilities from observed frequencies or from a known distribution.
Use the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The subtraction prevents double-counting outcomes that are in both A and B. If A and B are mutually exclusive (can't happen together), P(A ∩ B) = 0, and the formula simplifies to P(A) + P(B). Example: drawing a card that's a king or a heart from a deck. P(king) = 4/52, P(heart) = 13/52, P(king of hearts) = 1/52. So P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52.
For independent events, multiply: P(A ∩ B) = P(A) × P(B). Tossing two fair coins, P(heads on both) = 0.5 × 0.5 = 0.25. For dependent events, use the general multiplication rule: P(A ∩ B) = P(A) × P(B|A), where P(B|A) is the conditional probability. Example: drawing two aces from a deck without replacement. P(first ace) = 4/52, P(second ace given first was ace) = 3/51, so P(both aces) = (4/52)(3/51) ≈ 0.0045. Always check whether events are independent before reaching for the simple product.
Union (A ∪ B) is "A or B or both" — at least one of them happens. Intersection (A ∩ B) is "A and B" — both happen at the same time. In a Venn diagram, union is the total area covered by either circle; intersection is only the overlap in the middle. Numerically, P(A ∪ B) is usually larger than P(A ∩ B). Quick example: rolling a die where A = "even" and B = "greater than 3". A ∪ B = {2, 4, 5, 6}, A ∩ B = {4, 6}.
The formula is P(A|B) = P(A ∩ B) / P(B), where P(B) > 0. It tells you the probability of A given that B has already occurred. Example: in a class, 60% of students pass maths and 30% pass both maths and physics. P(physics pass | maths pass) = 0.30 / 0.60 = 0.50. Conditioning on B essentially shrinks the sample space to just the outcomes where B is true. Bayes' theorem builds directly on this idea, letting you update probabilities as new evidence arrives.
The complement of an event A is "A does not happen", written A' or Ac. Since one of A or A' must occur, P(A) + P(A') = 1, so P(A') = 1 − P(A). It's a powerful shortcut. Example: P(at least one head in 3 coin tosses). Computing this directly is fiddly, but the complement is "no heads at all" = (0.5)3 = 0.125. So P(at least one head) = 1 − 0.125 = 0.875. Always look for the complement when "at least one" or "not" appears in the problem.
Two events are independent when one happening doesn't change the probability of the other. For independent A and B, P(A ∩ B) = P(A) × P(B). Tossing a fair coin twice: P(heads then heads) = 0.5 × 0.5 = 0.25. Don't confuse independence with mutual exclusivity — those are opposite ideas. Mutually exclusive events can't happen together (P(A ∩ B) = 0), so they're actually highly dependent. Real-world independence often needs careful checking — drawing cards without replacement, for instance, makes successive draws dependent.
Convert your value to a z-score using z = (x − μ)/σ, then look up the area under the standard normal curve. Want P(X < 75) when μ = 70, σ = 10? Compute z = (75 − 70)/10 = 0.5, and the cumulative area at z = 0.5 is about 0.6915, so P(X < 75) ≈ 69.15%. For P(X > 75), take 1 − 0.6915 = 0.3085. Excel's NORM.S.DIST(z, TRUE) returns the cumulative probability. For a range, subtract two cumulative probabilities — that's the area between two z values.