Combining Probabilities, Explained

After reading this you can combine two independent events with AND, OR and XOR, find the chance of at least one success across many trials, and compute a conditional probability from a joint probability.

What this calculator does

Probability is a number between 0 and 1 that measures how likely an event is. This page handles three common questions that trip people up when they try to do the arithmetic in their heads.

Suppose you flip a fair coin and separately roll a die. The chance the coin lands heads is 0.5. The chance the die shows a six is about 0.1667. What is the chance both happen? What is the chance at least one happens? Those are two different questions with two different answers, and mixing them up is the single most common error in beginner probability.

The three modes cover:

Two independent events: given P(A) and P(B), find AND, OR and XOR.
At least one success in n trials: given a per-trial success probability p and a trial count n, find the chance of one or more successes.
Conditional probability: given P(A \cap B) and P(B), find P(A \mid B).

When independence applies, and when it does not

Two events are independent when knowing one occurred does not change the probability of the other. A coin flip and a die roll are independent. Two draws from a deck without replacement are not: after you draw an ace, the chance the next card is an ace drops from 4/52 to 3/51.

The two-events mode and the at-least-once mode both assume independence. If your events influence each other, the AND and OR formulas below give wrong answers. Use the conditional mode instead, or model the dependence directly.

The multiplication rule P(A \cap B) = P(A)\,P(B) only holds under independence. For dependent events the correct rule is P(A \cap B) = P(A)\,P(B \mid A). Assuming independence when it is false usually overstates the chance of a joint event.

The formulas and the intuition behind them

For two independent events with probabilities P(A) and P(B):

P(A \cap B) = P(A) \cdot P(B)

The AND probability multiplies because both conditions must hold. Each event thins out the other. With 0.5 and 0.3 you get 0.15, smaller than either input, as an intersection should be.

P(A \cup B) = P(A) + P(B) - P(A \cap B)

The OR probability adds the two, then subtracts the overlap once so it is not counted twice. Without the subtraction you would double count the region where both occur.

P(A \oplus B) = P(A) + P(B) - 2\,P(A \cap B)

XOR (exactly one, not both) subtracts the overlap twice: once to remove it from each of the two added terms. Here P(A) is the probability of A, P(B) the probability of B, and the cap and cup symbols mean intersection (both) and union (at least one).

For the at-least-once question, the trick is to work with the complement. The chance a single trial fails is 1 - p. The chance all n trials fail is that raised to the power n. So:

P(\text{at least one}) = 1 - (1 - p)^{n}

Conditional probability rescales. Once you know B happened, the whole sample space shrinks to B, so you divide the joint probability by P(B):

P(A \mid B) = \frac{P(A \cap B)}{P(B)}

A worked example with the demo numbers

Two events with P(A) = 0.5 and P(B) = 0.3

These are the calculator's default values. Load them and press Calculate to reproduce every number here.

AND: 0.5 × 0.3 = 0.15. Both events together happen 15% of the time.
OR: 0.5 + 0.3 - 0.15 = 0.65. At least one happens 65% of the time.
XOR: 0.5 + 0.3 - 2 × 0.15 = 0.50. Exactly one happens 50% of the time.
Check: XOR plus AND should equal OR. 0.50 + 0.15 = 0.65. It matches.

That last check is worth remembering. "At least one" (OR) splits cleanly into "exactly one" (XOR) plus "both" (AND).

At least one success with p = 0.1 and n = 10

Switch to the at-least-once mode and use the defaults p = 0.1, n = 10.

Failure per trial: 1 - 0.1 = 0.9.
All ten fail: 0.9^10 = 0.3487.
At least one succeeds: 1 - 0.3487 = 0.6513.

A 10% chance per attempt does not become 100% over ten attempts. It reaches about 65%. Multiplying 0.1 × 10 = 1.0 is the wrong instinct: probabilities do not add across trials like that.

How the at-least-once probability grows

The curve rises fast at first, then flattens as it approaches 1 but never reaches it. Each extra trial multiplies the failure probability by another 0.9, so the gap to certainty shrinks by 10% each step, not to zero.

With p = 0.1, ten trials reach about 0.65. You need 22 trials to pass 0.90 and 44 to pass 0.99.

Reading and interpreting the results

Every output is a probability between 0 and 1. Multiply by 100 for a percentage. A few sanity checks keep you honest:

AND is never larger than either input: Since you multiply two numbers each at most 1, P(A \cap B) \le \min(P(A), P(B)). If your AND result exceeds one of the inputs, something is wrong.
OR is never smaller than either input: P(A \cup B) \ge \max(P(A), P(B)), and never above 1.
Conditional can exceed both inputs: Dividing by P(B), a number below 1, can push P(A \mid B) above P(A \cap B). With P(A and B) = 0.2 and P(B) = 0.5, you get 0.2 / 0.5 = 0.4.

If independence holds, P(A \mid B) equals P(A), because knowing B tells you nothing about A. Comparing the conditional result to P(A) is a quick test of whether two events move together.

Common mistakes

Adding instead of the complement. For "at least one" people often add per-trial probabilities. With p = 0.1 and n = 10 that gives 1.0, which is nonsense. The right answer is 0.6513. If the sum ever exceeds 1, that alone proves the method is wrong.

Forgetting to subtract the overlap in OR. Writing P(A) + P(B) double counts the both-happen region. With 0.5 + 0.3 = 0.8 you overstate the true 0.65 by 0.15, exactly the AND value you skipped.

Assuming independence for dependent events. Card draws, sampling without replacement, and anything with a shared cause are dependent. Use the conditional formula.

Swapping the condition. P(A \mid B) and P(B \mid A) are different numbers unless the two events are equally likely. Reversing them is the classic base-rate error.

When a word problem says "given that," "among those who," or "of the ones that," you are almost always looking at a conditional probability. Identify what is being conditioned on before you compute.

Related calculators

To go from P(B \mid A) to P(A \mid B), including the diagnostic-test setup, use the Bayes' Theorem Calculator. When you need to count the arrangements behind a probability, the Combinations & Permutations Calculator gives nCr and nPr. For the average and spread of a random payoff, see the Expected Value Calculator. Two probability classics build directly on the at-least-once idea: the Dice Probability Calculator for sums across many dice, and the Birthday Paradox Calculator, which is a one-minus-complement calculation in disguise.

Frequently asked questions

What is the difference between OR and XOR?

OR (union) counts the both-happen case: at least one occurs. XOR counts only the exactly-one case. With P(A) = 0.5 and P(B) = 0.3, OR is 0.65 and XOR is 0.50. The difference, 0.15, is exactly the probability that both happen.

Why is the chance of at least one success not p times n?

Because probabilities do not add across independent trials. Multiplying 0.1 × 10 gives 1.0, implying certainty, which is false. The correct formula is 1 - (1 - p)^n = 0.6513. The product p × n is the expected number of successes, a different quantity.

Are P(A given B) and P(B given A) the same?

Only when P(A) = P(B). In general they differ. Converting between them requires Bayes' theorem, which uses the base rates of both events.

Can a probability be greater than 1?

No. Any valid probability sits between 0 and 1. If a calculation produces a number above 1, the method is wrong, most often an OR done without subtracting the overlap, or an at-least-once done by adding.

How do I find the chance that none of the trials succeed?

It is the complement of at least one: (1 - p)^{n}. With p = 0.1 and n = 10 that is 0.9^10 = 0.3487, and 0.3487 + 0.6513 = 1.

Try it yourself in the tool