Probability Calculator

Two probabilities and one question about how the events relate, and this returns the intersection, the union, both complements, the exclusive-or, and both conditional probabilities. Every one of them shows its formula with your numbers written into it, so you can copy the working and not just the answer. Enter probabilities however the question gave them to you: 0.25, 25%, 1/4 or 1 in 4 all work. A third event, repeated trials and Bayes' theorem are a click away underneath. Below, the and-versus-or split, the three relationships, a worked example, and conditional probability and Bayes.

P(A), the chance of event A

= 0.4

P(B), the chance of event B

= 0.3

How A and B relate

One happening tells you nothing about the other.

A or B, or both P(A ∪ B)

0.58

29/50 · 58%. The overlap P(A ∩ B) is 0.12, so it gets subtracted once to stop it being counted twice.

Each circle's area is its probability and the overlap is P(A ∩ B), so the picture changes as the numbers do. Mutually exclusive events pull apart until they only touch.

P(A ∩ B)0.12

Both A and B

P(A) × P(B)
= 0.4 × 0.3
= 0.12 = 3/25 = 12%

P(A ∪ B)0.58

A or B, or both

P(A) + P(B) − P(A ∩ B)
= 0.4 + 0.3 − 0.12
= 0.58 = 29/50 = 58%

P(A △ B)0.46

Exactly one of them

P(A) + P(B) − 2 × P(A ∩ B)
= 0.4 + 0.3 − 2 × 0.12
= 0.46 = 23/50 = 46%

P((A ∪ B)′)0.42

Neither one

1 − P(A ∪ B)
= 1 − 0.58
= 0.42 = 21/50 = 42%

P(A′)0.6

A does not happen

1 − P(A)
= 1 − 0.4
= 0.6 = 3/5 = 60%

P(B′)0.7

B does not happen

1 − P(B)
= 1 − 0.3
= 0.7 = 7/10 = 70%

P(A | B)0.4

A, given B happened

P(A ∩ B) ÷ P(B)
= 0.12 ÷ 0.3
= 0.4 = 2/5 = 40%

P(B | A)0.3

B, given A happened

P(A ∩ B) ÷ P(A)
= 0.12 ÷ 0.4
= 0.3 = 3/10 = 30%

Add a third event

P(C), the chance of event C

0.4 · 40% · 2/5 · 2 of 5

Enter P(C) above. C is taken as independent of both A and B, which is all two probabilities can tell you.

Repeat event A several times

How many trials

P(all n)0.000105

It happens every time

P(A)^n
= 0.4^10
= 0.000105 = 0.0105%

P(none)0.006047

It never happens

(1 − P(A))^n
= (1 − 0.4)^10
= 0.006047 = 0.6047%

P(≥ 1)0.993953

It happens at least once

1 − (1 − P(A))^n
= 1 − 0.6^10
= 0.993953 = 99.3953%

Bayes' theorem: flip a conditional probability

You know how often B follows A and want to know how often A was behind a B. The defaults are the medical testing example: a test that catches 99% of cases, a condition 1% of people have.

P(B | A)

= 0.99

P(A)

= 0.01

P(B)

= 0.0198

P(A | B)0.5

A, given B happened

P(B | A) × P(A) ÷ P(B)
= 0.99 × 0.01 ÷ 0.0198
= 0.5 = 1/2 = 50%

And, or, and the overlap in between

Almost every probability question turns on which of these four regions it is asking about. Taking P(A) = 0.4 and P(B) = 0.3 as independent events:

In words	Notation	Formula	Here
Both A and B	P(A ∩ B)	P(A) × P(B)	0.12
A or B, or both	P(A ∪ B)	P(A) + P(B) − P(A ∩ B)	0.58
Exactly one of them	P(A △ B)	P(A) + P(B) − 2 × P(A ∩ B)	0.46
Neither one	P((A ∪ B)′)	1 − P(A ∪ B)	0.42

The four add up to more than 1 only if you count wrong: the union already contains the intersection. The three regions that do partition everything are exactly-one, both, and neither, which come to 0.46 + 0.12 + 0.42 = 1.

Which of the three relationships is yours?

This is the choice that decides the answer, and it is where most marks are lost. Only the overlap changes, but every other value is built on it.

Independent

P(A ∩ B) = P(A) × P(B) = 0.12

One happening changes nothing about the other.

Two coin flips. Rain today and a coin landing heads.

Mutually exclusive

P(A ∩ B) = 0 = 0

They cannot both happen.

One die showing a 3 and the same die showing a 5.

Neither, so you need the overlap

P(A ∩ B) = the value you are given = 0.3

They affect each other, and independence would give the wrong answer.

Drawing two cards without putting the first one back.

Independent and mutually exclusive are opposites, not synonyms, though they are constantly swapped. Mutually exclusive events are about as dependent as two events can get: learning that one happened tells you the other definitely did not.

A worked example, start to finish

Two dice are rolled. A is "the first die shows a six", B is "the second die shows a six". Each is 1/6, and the dice do not influence each other, so they are independent.

Both sixes: P(A) × P(B) = 0.166667 × 0.166667 = 1/36
At least one six: P(A) + P(B) − P(A ∩ B) = 11/36, which is 30.5556%. Note it is not 1/6 + 1/6, because the double six would be counted twice.
Exactly one six: 5/18, the union minus the double.
Neither die a six: 25/36, which is (5/6)² and the quickest route to the answer in step 2.

Enter 1/6 in both boxes to see the same four numbers with the working attached.

Conditional probability and Bayes' theorem

P(A | B) asks what A's chances are once B is known to have happened. Knowing B shrinks the world down to B alone, so the overlap gets measured against that instead of against everything: P(A | B) = P(A ∩ B) ÷ P(B). With P(A) = 0.5, P(B) = 0.4 and an overlap of 0.3, that gives P(A | B) = 0.75 while P(B | A) = 0.6. The two directions are different numbers, and mixing them up has a name: the prosecutor's fallacy.

Bayes' theorem converts one direction into the other: P(A | B) = P(B | A) × P(A) ÷ P(B). A screening test that catches 99% of a condition 1% of people have, producing positives 1.98% of the time overall, means a positive result is a true positive 50% of the time. The test is not bad. The condition is just rare enough that false positives from the healthy 99% match the true positives one for one.

If A and B turn out to be independent, P(A | B) comes back equal to P(A), which is the cleanest definition of independence there is: the condition made no difference.

The same event, repeated

Repeat an event with probability p a total of n times, independently, and three questions have short answers. It happens every time: p^n. It never happens: (1 − p)^n. It happens at least once: 1 − (1 − p)^n, which is the complement of never and much easier than adding up every other case.

Rolling a fair die 10 times, a six comes up at least once 83.8494% of the time and never at all 16.1506% of the time. For a count of successes rather than a yes-or-no, the binomial distribution calculator handles exactly k, at least k and the rest.

How this calculator is checked

Every figure in the copy above, including the worked dice example and the Bayes result, is produced at build time by the same functions the calculator calls.
Impossible inputs are named rather than computed past: an overlap larger than either event, mutually exclusive probabilities summing over 1, and an overlap too small to keep the union under 1.
Conditioning on a zero-probability event returns undefined instead of NaN or zero, so a division by zero is never presented as an answer.
The Venn diagram is drawn from the numbers, with circle areas proportional to P(A) and P(B) and the overlap solved to match P(A ∩ B), rather than being a fixed illustration.
Everything runs in your browser. Nothing you enter is stored or sent to a server.

Last reviewed: October 2026. Found an error? Report it and we'll fix it.

Frequently Asked Questions

What is the difference between P(A and B) and P(A or B)?

And is the overlap: both events happen. Or is the union: at least one happens, which includes the case where both do. For independent events with P(A) = 0.4 and P(B) = 0.3, the and is 0.12 and the or is 0.58. The or is always the larger of the two, because the overlap is part of it.

Why do you subtract P(A and B) from P(A) + P(B)?

Because adding the two probabilities counts the overlap twice, once inside P(A) and once inside P(B). Subtracting it once puts it back to a single count. That is the addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). When the events are mutually exclusive the overlap is zero and the subtraction does nothing, which is why that case looks like plain addition.

When can I just multiply two probabilities?

Only when the events are independent, meaning one happening tells you nothing about the other. Multiplying dependent events is the single most common error in this topic. Drawing two aces from a deck without replacement is 4/52 × 3/51, not 4/52 × 4/52, because the first draw changed what was left. Use the third option in the tool and enter the real overlap when independence does not hold.

What is conditional probability?

P(A | B) is the chance of A once you already know B happened. It is the overlap divided by the thing you now know: P(A ∩ B) ÷ P(B). Knowing B shrinks the world to just B, so the overlap is measured against that smaller world instead of against everything. If A and B are independent, P(A | B) equals P(A), because knowing B taught you nothing.

How do you calculate the probability of 3 events?

For three independent events, all three happening is P(A) × P(B) × P(C). At least one happening is easier through the complement: 1 − (1 − P(A))(1 − P(B))(1 − P(C)), because none of them happening is a single product. Open the third-event section in the calculator and it shows all three happening, at least one, exactly one, and none.

What is the probability of something happening at least once in n tries?

1 − (1 − p)^n. Rolling a six at least once in 10 rolls of a fair die is 1 − (5/6)^10 = 0.838494, or 83.8494%. Note it never reaches 1: more tries push it closer to certainty without ever arriving. The repeated-trials section computes this along with always and never.

What is Bayes' theorem used for?

Flipping a conditional around when you know one direction and want the other. The classic case is medical testing: a test that catches 99% of a condition 1% of people have, with a 1.98% overall positive rate, means a positive result is right only 50% of the time. The result is surprising because it depends on how rare the condition is, not just on how good the test is.

Can I enter probabilities as fractions or percentages?

Yes. Every probability box takes 0.25, 25%, 1/4, or "1 in 4", and shows what it read the entry as underneath. Answers come back as a decimal, as an exact fraction where one exists, and as a percentage, so you can copy whichever form the question asked for.

Why does the calculator warn me about my numbers?

Because some combinations describe something that cannot exist. An overlap larger than either event on its own, two mutually exclusive events whose probabilities add to more than 1, or an overlap so small that P(A ∪ B) would exceed 1. It computes the result anyway and tells you which input is the impossible one, rather than silently returning a number you would copy down.

What is P(A | B) when P(B) is zero?

Undefined. You would be dividing by zero, and conditioning on something that never happens has no meaning. The calculator says undefined rather than showing NaN or 0, both of which read like answers.

For dice specifically, the dice probability calculator handles totals, ranges and matching dice, and draws the whole distribution for the dice you rolled. For a fixed number of independent trials at one success rate, the binomial distribution calculator gives P(X = k) and every tail.

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