Probability Calculator Calculate the probability of events, unions, intersections, normal distributions, and more.
Modify the values and click Calculate to get instant results.
i
Modify the values and click the Calculate button to use
Probability of Two Events
Find union, intersection, and related probabilities of two independent events.
Probability of A: P(A)
Probability of B: P(B)
Please input values between 0 and 1.
Result
P(A′) — A NOT occurring
P(B′) — B NOT occurring
P(A∩B) — Both A and B occur
P(A∪B) — A or B or both occur
P(A▵B) — A or B but NOT both
P((A∪B)′) — Neither A nor B
Probability Solver for Two Events
Provide any 2 values below to calculate the rest (independent events).
P(A)
P(B)
P(A′) — A NOT occurring
P(B′) — B NOT occurring
P(A∩B) — Both occur
P(A∪B) — A or B or both
P(A▵B) — A or B, not both
P((A∪B)′) — Neither
Please input values between 0 and 1.
Result
P(A)
P(B)
P(A′)
P(B′)
P(A∩B)
P(A∪B)
P(A▵B)
P((A∪B)′)
Probability of a Series of Independent Events
ProbabilityRepeat Times
Event A
Event B
Result
P(Event A repeated n times)
P(Event B repeated n times)
Combined Probability (all events)
Probability of a Normal Distribution
Find the area P under a normal distribution curve and confidence intervals.
Mean (μ)
Std Deviation (σ)
Left Bound (Lb)
Use -inf for −∞
Right Bound (Rb)
Use inf for +∞
Result
Conditional Probability (Bayes' Theorem)
Calculate P(A|B) using Bayes' Theorem: P(A|B) = P(B|A) × P(A) / P(B)
Probability of A: P(A)
Probability of B: P(B)
P(B|A) — B given A
Result
P(A|B) — Probability of A given B
P(A′|B) — Probability of NOT A given B
P(A|B′) — Probability of A given NOT B
Permutations & Combinations
Calculate permutations (nPr) and combinations (nCr) for given n and r values.
Total items (n)
Items chosen (r)
Result
Permutations — nPr = n! / (n−r)!
Combinations — nCr = n! / (r!(n−r)!)
User Guide

How to Use:

  1. Select the calculator type you need
  2. Enter your probability values (between 0 and 1)
  3. Click Calculate to see results
  4. Export results as PDF using the Export button

Understanding Results:

  • P(A′) — Complement (event NOT occurring)
  • P(A∩B) — Intersection (both events occur)
  • P(A∪B) — Union (either or both occur)
  • P(A▵B) — Exclusive OR (one but not both)
  • P(A|B) — Conditional (A given B)

Tips:

  • All probability values must be between 0 and 1
  • The Solver needs at least 2 values to compute the rest
  • Use -inf and inf for infinite bounds in normal distribution
  • For Bayes' theorem, P(B) must not be zero
Formulas Reference

1. Complement:
P(A′) = 1 − P(A)

2. Intersection (independent):
P(A∩B) = P(A) × P(B)

3. Union:
P(A∪B) = P(A) + P(B) − P(A∩B)

4. Exclusive OR:
P(A▵B) = P(A∪B) − P(A∩B)

5. Series:
P = P(A)n × P(B)m × ...

6. Bayes' Theorem:
P(A|B) = P(B|A) × P(A) / P(B)

7. Normal Distribution:
P(Lb < X < Rb) = Φ((Rb−μ)/σ) − Φ((Lb−μ)/σ)

8. Permutations:
nPr = n! / (n−r)!

9. Combinations:
nCr = n! / (r!(n−r)!)

10. Confidence Interval:
CI = μ ± z × σ

Probability Calculator: Work Out the Odds of Almost Anything

With this probability calculator you can find out the odds of practically anything without having to know the statistical name for your situation first. So probability is not one calculation. It is a bunch of distinct calculations depending on the nature of the question that you’re actually asking. This page is set up just like that. It is organized by what your query looks like, not by textbook chapter names. So you can locate the proper solution even if you’ve never heard the phrase “binomial” before. We’ll work through single events, mixed events, dice and coins, specifically repeated trials, and continuous measures, all with a fully functional example you can check step by step against your own figures.

One Event: The Basic Probability Formula

This is the most fundamental form any probability inquiry may take: what are the odds of one particular result against all that could happen?

P(A) = no of good outcomes / total no of potential outcomes

Worked example. What is the probability of obtaining a red card from a regular 52-card deck?

There are 26 red cards (hearts and diamonds) out of 52 cards altogether.

P(red) = 26/52 = 0.5 or 50%

Here is another useful rule, which silently solves a substantial share of genuine questions: the complement rule. The odds of something not happening are 1 minus the odds of it happening.

P(not A) = 1 – P(A)

For our card example, P(not red) = 1 – 0.5 = 0.5 or 50%. The use of this probability calculator makes sense here as the deck is evenly split. This rule is quite handy later on, where computing “at least one” directly would be much more complicated than computing its inverse.

Two or More Events: Where Most People Go Wrong

This is the question type that creates the most errors, and the error nearly always boils down to one distinction: are the events independent (the outcome of one doesn’t affect the outcome of the next) or dependent (it does)?

For independent occurrences, we can use the multiplication rule directly: P(A and B) = P(A) P(B).

Worked example: What’s the chance of flipping two coins and getting heads both times?

P (heads and heads) = 0.5 * 0.5 = 0.25, 25 %

Each flip does not affect the other; hence, this is a clean multiplication.

Dependent events involve a shift: the outcome of the first event influences the probable outcomes of the second event. P (A and B) = P (A) x P (B | A)

Worked example: What is the likelihood of drawing two aces in a succession from a regular deck without replacing the first card?

First ace = 4/52. If you take one of the aces out, then you have 51 cards left, and only 3 of those are aces: P(second Ace|first was an Ace) = 3/51

P(both aces) = (4/52)*(3/51) = 12/2652 ~= 0.0045 or ~0.45%

The problem to see plainly here is that if you considered these draws as independent (forgetting that the first card was eliminated), you’d get (4/52) x (4/52) = 16/2704 = 0.59%, much higher than the accurate 0.45%. But that gap, however little it might seem, is precisely the sort of inaccuracy that arises from failing to distinguish between independent and dependent, and it is the single most prevalent source of inaccurate answers to this kind of question.

Also, another rule for compound events: if you want the probability of either one event or another event happening, you can use the addition rule: P(A or B) = P(A) + P(B) − P(A and B).

Worked example: What is the likelihood of drawing a king or a heart from a deck?

P (king) = 4/52, P (heart) = 13/52, and P (king and heart, i.e., the king of hearts) = 1/52.

P(king or heart) = 4/52 + 13/52 − 1/52 = 16/52 ≈ 30.8%

That subtraction is important, since without it we would count the king of hearts twice, once as a king and once as a heart, giving the answer as 17/52 instead of the correct 16/52.

Dice and Coin Probability

The most common serious questions people seek answers for are dice probability calculators, and coins, and they need meaningful treatment, not to be shrugged off as inconsequential.

One die, landing on a certain number: each face has a 1 in 6 probability of being rolled. P(rolling a 6) = 1/6 = 16.7%

Two dice, rolling a specific total: This one trips folks up because not every sum is equally likely. When you roll two dice there are 36 different combinations (6 x 6), but only some totals can be made multiple ways. Of 7 total advantageous combinations, 6 are, and they are (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).

P(total of 7) = 6/36 = 1/6, or about 16.7%, making 7 the most likely total on two dice.

Here’s the full reference table of two-dice totals:

Total

Combinations

Probability

2

1

2.80%

3

2

5.60%

4

3

8.30%

5

4

11.10%

6

5

13.90%

7

6

16.70%

8

5

13.90%

9

4

11.10%

10

3

8.30%

11

2

5.60%

12

1

2.80%

A 6 in at least 4 tosses of a die: This is when the earlier complement rule saves real work. To calculate “at least one 6” directly, we would need to add the probabilities of getting exactly one 6, exactly two 6s, exactly three 6s, and exactly four 6s separately. Much easier to calculate the inverse, “no 6s at all.”

P(6 in one roll) = 5/6. P(no 6 in four rolls) = (5/6)⁴ ~ 0.4823. This is for four independent rolls.

P(at least one 6) = 1 – 0.4823 = 0.518 ~ 51.8%

So if you roll a die 4 times, you have slightly better than even odds of rolling a 6 at least once. This surprises a lot of people who think it should be closer to 4 × (1/6) = 67%. That’s a common mistake, thus worth flagging up directly: you can’t just add individual probabilities together in that way when there’s a possibility of the same thing happening more than once during a series of trials.

Repeated Trials: Binomial Probability

It is called a binomial situation when the same trial happens a set number of times and each time has the same chance of turning out well.

P(X = k) = C(n,k) × pᵏ × (1 – p)ⁿ⁻ᵏ

C(n,k) is the number of ways to pick k successes from n trials. n is the number of trials, k is the number of successes you want, and p is the chance of success on a single trial.

As an example, what is the chance of getting exactly three heads out of five flips of the coin?

There are 5 n, 3 k and a p-value of 0.5.

When you flip 5 coins, 3 heads and 2 tails can show up in 10 different ways. This is called C(5,3).

P(X = 3) = 10 x 0.5³ x 0.5² = 10 x 0.125 x 0.25 = 10 x 0.03125, or 31.25%

By separating each part in this way, you can do the math again with your own numbers instead of leaving the formula as a mystery. This method works for “at least k successes” questions too; it’s called a cumulative binomial probability calculator, and it works by adding up the odds for k, k+1, k+2, and so on up to n. The probability calculator above takes care of this for you.

Measurements and Ranges: Normal Probability

If you are working with continuous data that clusters around an average (heights, test scores, measurement mistakes, etc.), then the normal distribution (or the classic bell curve) is usually the perfect tool.

The z-score is the link between a raw measurement and a probability:

z = (value – mean) / standard deviation

A z-score tells you how many standard deviations a given number is from the mean. This can then be translated into a probability calculator based on the features of the standard normal probability calculator curve.

Worked example. A data collection can have a mean of 100 and a standard deviation of 15. What is the likelihood that a randomly selected value is less than 115?

z = (115 – 100) / 15 = 1.0

A z-score of 1.0 equates to about 84 percent on the standard normal distribution. (Source: Standard normal distribution table, an established statistical reference) In simple terms, 115 is 1 standard deviation above the mean. The bell curve has most of its area below the mean; therefore, around 84% of all the values in this dataset will be below 115.

A Note on Probability of Default

If you were looking for information on calculating probability of default, this is a specialized notion within credit risk and not a general probability calculator use, which is covered on this site.

Probability of Default (PD) is the estimated probability that a borrower would default on their debt obligations over a specified period of time and is a criterion used by lenders and in the modeling of bond risk. The building blocks of credible PD estimations are institution-specific historical default data, credit ratings, and macro economic inputs. These are incorporated in models that are significantly more specialized than a general probability formula. This is really outside the purview of a general probability calculator, and this page is not going to provide an alternative formula that is likely to be confused with a usable credit risk calculation. If you are doing real credit risk modeling, then that is a job for institution-specific financial risk resources, not a general-purpose probability calculator.

Conclusion

Most problems about probability calculators fall into one of four shapes: a single event, combined occurrences, repeated trials, or a continuous measurement, and each has a simple technique once you know which form you’re dealing with. The independent versus dependent distinction described above is the cause of more wrong answers than any other single mistake. So it’s worth checking carefully whenever you’re combining two or more events, and it’s the one detail that’s worth re-reading if a real-world answer ever comes out looking surprisingly different from what intuition first suggested.

FAQs

Q1. How do you calculate probability?

Take the number of favorable results and divide it by the number of total possible outcomes. That is 26 favorable occurrences out of 52 overall, yielding 26/52 = 0.5 or 50% for drawing a red card from a regular deck.

Q2. What’s the difference between independent and dependent events?

Independent occurrences are events that do not effect each other’s outcome, like different flips of a coin. Dependent events do. Such as drawing cards without replacement. Confusing dependent events with independent events, like in our case of the two aces, increased the answer from the accurate 0.45% to an incorrect 0.59%.

Q3. What are the odds of rolling a 7 with two dice?

16.7%, 1 in 6. There are 36 possible combinations on two dice and 6 combinations to reach a total of 7. This is more than any other total and makes 7 the single most frequent result.

Q4. When do I use binomial probability?

Like flipping a coin 5 times, or seeing if a fixed number of products pass a quality test. When the same trial is repeated a fixed number of times and each has the same likelihood of success. In our worked example, we found the probability calculator of getting exactly 3 heads in 5 tosses of a coin was 31.25%.

Q5. What is a z-score?

A statistic that shows how many standard deviations the value is from its mean: (value – mean) / standard deviation. For example, a z score of 1.0 in our worked example means that around 84% of values are below this point.

Q6. What’s the difference between probability and odds?

Probability is a way of describing the likelihood that an event might happen. It is usually written as a fraction of all the potential outcomes. For example, the probability of getting a given number when you roll a die is 1/6. Odds say it’s a ratio of favorable occurrences to unfavorable outcomes, like 1-to-5 for the same dice roll. Both describe the same underlying chance, albeit in different formats.