Probability Calculator

Find the chance of two events happening together, either one, or neither, or the chance of a number of successes in repeated tries, with every step shown.

Solve for

The chance that event A happens, as a decimal (0.25) or percent (25%).

The chance that event B happens.

A and B are

P(A and B)

15.0000%

P(A or B) = 65.0000% · independent events

  • P(A and B)0.15 · 15.0000%
  • P(A or B)0.65 · 65.0000%
  • P(exactly one of A, B) — one happens, not both50.0000%
  • P(A only) · P(B only)35.0000% · 15.0000%
  • P(neither) — neither happens35.0000%
  • P(not A) · P(not B)50.0000% · 70.0000%
  • P(A | B) — chance of A if B happened50.0000%
  • Odds of A and B1 in 6.67

How this was calculated

  1. “A and B” means both happen. The events are independent (one does not change the chance of the other), so multiply their chances.P(A and B) = P(A) × P(B)Put in your numbers: 0.5 × 0.3Answer: 0.15 (15.0000%)
  2. “A or B” means at least one of them happens. Add the two chances, then take away the overlap so it is not counted twice.P(A or B) = P(A) + P(B) − P(A and B)Put in your numbers: 0.5 + 0.3 − 0.15Answer: 0.65 (65.0000%)
  3. The chance that something does not happen is 1 minus the chance that it does.P(not A) = 1 − P(A) and P(not B) = 1 − P(B)Put in your numbers: 1 − 0.5 and 1 − 0.3Answer: 50.0000% and 70.0000%
  4. “Neither” is the opposite of “A or B”: take it away from 1.P(neither) = 1 − P(A or B)Put in your numbers: 1 − 0.65Answer: 35.0000%
  5. “Exactly one” means A or B happens, but not both.P(exactly one) = P(A or B) − P(A and B)Put in your numbers: 0.65 − 0.15Answer: 50.0000%
  6. P(A | B) is the chance of A when you already know B happened. For independent events it is the same as P(A).P(A | B) = P(A and B) ÷ P(B)Put in your numbers: 0.15 ÷ 0.3Answer: 50.0000%

Probability in simple words

Probability is how likely something is to happen, as a number from 0 to 1. 0 means it can never happen, 1 means it is certain, and 0.5 means it is an even chance, like a coin landing on heads. You can also write it as a percent: 0.5 is 50%.

When two things do not affect each other (they are independent), the chance of both happening is the two chances multiplied. A coin landing heads (0.5) and a die showing 6 (1/6 ≈ 0.1667) together is 0.5 × 0.1667 ≈ 0.083, about 8%.

The chance of one or the other (or both) is the two chances added, minus the overlap so it is not counted twice: 0.5 + 0.1667 − 0.083 ≈ 0.583, about 58%.

The chance that something does not happen is 1 minus the chance that it does. The die not showing 6 is 1 − 1/6 = 5/6, about 83%.

Two events: independence versus exclusion

For independent events, P(A and B) = P(A)P(B). For mutually exclusive events, P(A and B) = 0. In both cases, P(A or B) = P(A) + P(B) − P(A and B). Independence and mutual exclusion are different assumptions, not interchangeable options.

For P(A) = 0.5 and P(B) = 0.3 with independence, the chance of both is 0.15, either is 0.65, and neither is 0.35. General dependent events require additional information and are not modeled here.

Exactly, at least and at most

For n independent trials with constant success probability p, P(X = k) = C(n, k)p^k(1 − p)^(n − k). Add terms from k through n for “at least k”, or from 0 through k for “at most k”. Expected successes equal np, and standard deviation is √[np(1 − p)].

In ten fair coin flips, exactly three heads has probability 120 ÷ 1,024 = 0.1171875. At most three heads is 0.171875; at least three is 0.9453125. The chance of at least one head is 1 − (0.5)^10 = 0.9990234375.

Input and model limits

Enter a probability as a decimal from 0 to 1 or with an explicit percent sign, such as 25%. A bare 25 is rejected. Trials must be whole numbers from 1 to 1,000 and successes must be whole numbers between zero and the trial count.

Repeated-trial models assume the same success probability on each independent trial. Sampling without replacement generally violates that assumption. Very small displayed probabilities can round to zero; logarithmic computation reduces loss from intermediate powers but cannot represent arbitrarily tiny values.

Frequently asked questions

Why is at least one calculated using the complement?

The opposite of at least one success is no successes. Independence gives P(none) = (1 − p)^n, so P(at least one) = 1 − (1 − p)^n.

Can mutually exclusive events have probabilities adding above 1?

No. Their union is the sum of their probabilities, and any probability must be at most 1. The calculator rejects such a combination.

Last updated . Results are estimates for informational purposes only.