Probability of Multiple Events Calculator

The probability that several events happen depends on how the events relate to each other, so the calculator has three modes:

• Two events: A and B. Enter P(A) and P(B), then choose whether the events are independent (one does not affect the other), mutually exclusive (they cannot both happen) or overlap by a probability P(A and B) that you know. Every other probability of the pair follows from these three numbers.
• Several independent events. Enter one probability per event, for as many as 50 events. You get the chance that all of them happen, that none does, that at least one does and that exactly k do, for every k.
• Dependent events, one after another. Enter P(A₁), then P(A₂ | A₁), then P(A₃ | A₁ and A₂) and so on. Their product is the chance that all the events happen, as when cards are dealt without putting them back.

A probability can be typed as a decimal (0.25), a percentage (25%) or a fraction (1/4). The calculator keeps it as an exact fraction, so every sum, product and difference is exact and nothing is rounded until the answer is written.

The “or” of probability is inclusive: P(A or B) is the probability that A happens, B happens or both do. Adding P(A) and P(B) counts the outcomes where both happen twice, so P(A and B) is subtracted once. For mutually exclusive events that overlap is 0 and nothing is subtracted. For “at least one” it is easier to find the opposite, that none happens, and subtract it from 1.

Suppose P(A) = 0.4 and P(B) = 0.5, and the events do not affect each other. Load example fills in these values.

1. P(A and B) = 0.4 × 0.5 = 0.2.
2. P(A or B) = 0.4 + 0.5 − 0.2 = 0.7. Adding the two probabilities alone would give 0.9, which counts the overlap twice.
3. P(A only) = 0.4 − 0.2 = 0.2 and P(B only) = 0.5 − 0.2 = 0.3, so P(exactly one) = 0.5.
4. P(neither) = 1 − 0.7 = 0.3, which is also 0.6 × 0.5, the product of the two complements.
5. Independence: P(A | B) = 0.2 / 0.5 = 0.4, the same as P(A), so knowing that B happened does not change the probability of A.

Draw one card from a standard deck of 52. Let A be “the card is a king”, so P(A) = 4 / 52 = 1/13, and B “the card is a heart”, so P(B) = 13 / 52 = 1/4. The king of hearts is in both events, so P(A and B) = 1/52. Choose I know P(A and B) and enter 1/13, 1/4 and 1/52.

1. P(A or B) = 1/13 + 1/4 − 1/52 = 16/52 = 4/13 ≈ 0.307692: there are 16 cards that are a king, a heart or both (4 kings and 13 hearts, less the king of hearts counted twice).
2. P(A only) = 3/52 ≈ 0.057692 (the three kings that are not hearts) and P(B only) = 12/52 = 3/13 ≈ 0.230769 (the twelve hearts that are not kings).
3. P(neither) = 1 − 4/13 = 9/13 ≈ 0.692308.
4. Independence: P(A) × P(B) = 1/13 × 1/4 = 1/52 = P(A and B), so Independent events? answers Yes. The suit of a card tells you nothing about its rank.

Draw one card and let A be “a king” and B “a queen”. A card cannot be both, so the events are mutually exclusive: choose Mutually exclusive and enter 1/13 for both.

1. P(A and B) = 0.
2. P(A or B) = 1/13 + 1/13 = 2/13 ≈ 0.153846, the 8 kings and queens out of 52 cards.
3. Independence: P(A) × P(B) = 1/169 ≈ 0.005917, which is not P(A and B) = 0, so the events are not independent. After a king is drawn a queen is impossible, and the conditional probability P(B | A) is 0, not 1/13.

Three machines break down independently on a given day with probabilities 0.1, 0.2 and 0.05. Choose Several independent events and enter 0.1, 0.2, 0.05 (or use Load example).

1. P(all events happen) = 0.1 × 0.2 × 0.05 = 0.001: all three break down together once in a thousand days.
2. P(none happens) = 0.9 × 0.8 × 0.95 = 0.684.
3. P(at least one happens) = 1 − 0.684 = 0.316: on 31.6% of days at least one machine is down.
4. P(exactly one happens) = 0.1 × 0.8 × 0.95 + 0.9 × 0.2 × 0.95 + 0.9 × 0.8 × 0.05 = 0.076 + 0.171 + 0.036 = 0.283. The table adds P(exactly 2) = 0.032, and the four probabilities 0.684, 0.283, 0.032 and 0.001 add up to 1.
5. Expected number of events = 0.1 + 0.2 + 0.05 = 0.35 machines down on an average day.

Two cards are dealt from a deck of 52 without putting the first back. What is the chance that both are aces? The first card is an ace with probability 4/52. Given that, 3 aces are left among 51 cards, so the second is an ace with probability 3/51. Choose Dependent events, one after another and enter 4/52, 3/51.

1. P(all events happen) = 4/52 × 3/51 = 1/221 ≈ 0.004525, about 0.45%.
2. Continuing with 2/50 and 1/49 gives the chance of being dealt all four aces from the first four cards: 1/270725 ≈ 3.6938e-6.
3. If the first card were put back each time, the draws would be independent and the chance of two aces would be 1/13 × 1/13 = 1/169 ≈ 0.005917. Treating dependent draws as independent overstates the chance here.

• Independent events do not affect each other: knowing that one happened leaves the probability of the other unchanged. Coin flips, dice rolls, machines in different places and draws with replacement are independent. Then P(A and B) = P(A) · P(B).
• Mutually exclusive events cannot happen together, such as a king and a queen on one draw. Then P(A and B) = 0 and P(A or B) = P(A) + P(B). Two mutually exclusive events that can each happen are always dependent, because once one has happened the other is impossible.
• Dependent events change each other's probability, as when cards are dealt without replacement. Use P(B | A) for the second event. The conditional probability calculator finds it from counts or from P(A and B).

If you know only P(A) and P(B), the probability P(A and B) is not fixed by them: it can be anything from max(0, P(A) + P(B) − 1) to min(P(A), P(B)). For P(A) = 0.7 and P(B) = 0.6 that is from 0.3 to 0.6, which is why the calculator refuses a P(A and B) of 0.2 for these two events: it would make P(A or B) larger than 1.

For the two ideas side by side, with the tests and worked examples, see independent vs mutually exclusive events.

• Adding for “and”. The chance that two independent events both happen is the product of their probabilities, not the sum: 0.4 × 0.5 = 0.2, not 0.9.
• Adding for “or” without removing the overlap. P(A) + P(B) is right only for mutually exclusive events. Otherwise subtract P(A and B), as in 0.4 + 0.5 − 0.2 = 0.7.
• Adding for “at least one”. Three independent events with probability 0.1 each do not give 0.3. The chance that at least one happens is 1 − 0.9 × 0.9 × 0.9 = 0.271. Use the list mode or the at least one probability calculator for identical tries.
• Treating dependent events as independent. Drawing two aces without replacement has probability 1/221, not 1/169. Use the third mode and enter the conditional probability of each step.
• Confusing mutually exclusive with independent. They are nearly opposites: exclusive events with positive probabilities are dependent.
• Rounding an input. A third is 1/3, not 0.33. Type the fraction and the result stays exact.

These tools use binary floating-point numbers. For p = [0.1, 0.2, 0.05] Python prints the product as 0.0010000000000000002 and the chance of at least one as 0.31599999999999995. This calculator works with exact fractions, so it gives exactly 0.001 and 0.316. To count outcomes rather than multiply probabilities, use the combination calculator or the permutation calculator; for the chance of k successes in n identical independent tries, see the binomial distribution calculator, and for draws without replacement from a group of two kinds, the hypergeometric distribution calculator.