Probability Calculator with Combinations & Permutations

Calculate the probability of any event, combination, or permutation instantly. Get step-by-step formulas, decimal and percentage results, and real worked examples — all for free.

How to Use This Probability Calculator

Follow these three steps to get accurate probability results instantly:

1

Select Calculation Type

Choose Basic Probability to find event likelihood, Combinations when order doesn't matter (choosing a team), or Permutations when order does matter (ranking finalists).

2

Enter Your Values

For basic probability, enter your favorable outcomes and total outcomes. For combinations or permutations, enter n (total items) and r (items selected or arranged).

3

Read Your Results

Click Calculate Probability to see your result as both a decimal and a percentage, plus the exact formula used. Scroll down for worked examples to verify your understanding.

The Formula Explained

Learn how each probability formula works — and see it applied to real numbers you can verify yourself in the calculator above.

Basic Probability: P(E) = Favorable ÷ Total

The most fundamental formula in probability states that the likelihood of an event equals the number of ways it can occur divided by the total number of equally likely outcomes. This assumes every outcome in your sample space has the same chance of happening — a fair die, a well-shuffled deck, an unbiased coin.

Worked Example: Marble Drawing

A bag contains 4 red marbles, 3 blue marbles, and 5 green marbles (12 total). What is the probability of drawing a blue marble at random?

  • • Favorable outcomes: 3 (blue marbles)
  • • Total outcomes: 12 (all marbles in the bag)
  • • P(Blue) = 3 ÷ 12 = 0.25 (25%)

Try it: Enter Favorable = 3, Total = 12 in Basic Probability mode above.

The result tells you that on any single random draw, you have a 1-in-4 chance of pulling a blue marble. Repeat the experiment 100 times and you'd expect roughly 25 blue draws — though the exact number varies due to random variation.

Combinations: C(n, r) = n! ÷ (r! × (n − r)!)

Use combinations when you're choosing a group of items and order does not matter. The exclamation mark (!) means factorial — the product of all positive integers up to that number. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Since order is irrelevant, choosing {Alice, Bob, Carol} is the same combination as {Bob, Carol, Alice}.

Worked Example: Forming a Project Committee

Your department has 10 employees. How many different 3-person committees can you form?

  • • n = 10 (total employees), r = 3 (committee size)
  • • C(10, 3) = 10! ÷ (3! × 7!) = 3,628,800 ÷ (6 × 5,040) = 3,628,800 ÷ 30,240 = 120 possible committees

Try it: Select Combinations, enter n = 10, r = 3, then click Calculate.

Combinations appear whenever selection without ranking is involved: choosing lottery numbers, picking a jury, assembling a playlist, or selecting items in a grocery order. Any time you ask "how many groups?" — use C(n, r).

Permutations: P(n, r) = n! ÷ (n − r)!

Use permutations when the order of selection does matter. Ranking athletes for gold, silver, and bronze is a permutation problem — "Alice 1st, Bob 2nd" is a different event outcome than "Bob 1st, Alice 2nd." Because more arrangements are possible than groups, P(n, r) is always greater than or equal to C(n, r).

Worked Example: Science Fair Medals

At a science fair with 8 contestants, how many different ways can gold, silver, and bronze medals be awarded?

  • • n = 8 (contestants), r = 3 (medal positions)
  • • P(8, 3) = 8! ÷ (8 − 3)! = 40,320 ÷ 120 = 336 possible arrangements
  • • Compare: C(8, 3) = only 56 unordered groups of 3 — order multiplies the count by 6 (= 3!)

Try it: Select Permutations, enter n = 8, r = 3, then click Calculate.

Conditional Probability: P(A | B) = P(A ∩ B) ÷ P(B)

Conditional probability measures the statistical likelihood of event A occurring given that event B has already happened. The vertical bar "|" is read as "given that." This formula is the engine behind medical diagnosis, fraud detection, spam filters, and risk models throughout the US financial industry.

Worked Example: Playing Card Conditional Probability

From a standard 52-card deck, what is the probability that the drawn card is the Ace of Spades, given you already know it is a spade?

  • • P(Ace ∩ Spade) = 1/52 — only one Ace of Spades in the deck
  • • P(Spade) = 13/52 = 0.25 — there are 13 spades total
  • • P(Ace | Spade) = (1/52) ÷ (13/52) = 1/13 ≈ 0.0769 (7.69%)

Notice how the condition "it is a spade" narrows the sample space from 52 cards to 13 cards, making the probability jump from 1/52 ≈ 1.9% to 1/13 ≈ 7.7%.

Common Use Cases & Tips

From classrooms to boardrooms, probability calculations appear in every field. Here are six real-world scenarios with exact numbers that show you how to apply the formulas above.

1. Calculating Lottery Odds

Powerball requires you to match 5 white balls drawn from 1–69 and 1 red Powerball drawn from 1–26. Because the order you match the white balls doesn't matter, you use the combination formula:

  • • White balls: C(69, 5) = 11,238,513 possible combinations
  • • Red Powerball: C(26, 1) = 26 possibilities
  • • Jackpot odds: 11,238,513 × 26 = 1 in 292,201,338

Tip: Enter n = 69, r = 5 in Combinations mode, then multiply the result by 26 manually to confirm the jackpot odds.

2. Quality Control in Manufacturing

A production line makes 500 circuit boards per shift and quality checks find 15 defective units. You want to know the probability of randomly selecting a defective board, and the probability of selecting two defective boards in a row (without replacement):

  • • P(1st board defective) = 15 ÷ 500 = 0.03 (3%)
  • • P(2nd defective | 1st was defective) = 14 ÷ 499 ≈ 0.0281
  • • P(both defective) = 0.03 × 0.0281 ≈ 0.000843 (0.084%)

Since the first board is removed before the second draw, these are dependent events — the denominator drops from 500 to 499.

3. Multiple-Choice Exam Guessing

On a 5-choice standardized test question, the probability of guessing correctly at random is 1 in 5. What happens over a 10-question section?

  • • P(1 correct guess) = 1 ÷ 5 = 0.20 (20%)
  • • P(3 consecutive correct guesses) = 0.20 × 0.20 × 0.20 = 0.008 (0.8%)
  • • P(at least one wrong in 10 questions) = 1 − (0.20)¹⁰ ≈ 99.9999%

Use the complement rule: instead of adding up probabilities of getting exactly 1, 2, 3… wrong, just subtract P(all correct) from 1.

4. Sports Analytics & Betting Odds

In baseball, a batter with a .300 average has a 30% chance of getting a hit in any given at-bat (assuming independent events). A scout or analyst might ask: what's the probability of hitting safely in at least 1 of their next 4 at-bats?

  • • P(hit in one at-bat) = 0.300
  • • P(out in one at-bat) = 1 − 0.300 = 0.700
  • • P(no hits in 4 at-bats) = (0.70)⁴ = 0.2401
  • • P(at least one hit in 4 at-bats) = 1 − 0.2401 = 0.7599 (76%)

The complement rule makes this far easier than summing individual hit scenarios.

5. Medical Test Accuracy & Bayes' Theorem

A disease affects 1% of the US population. A screening test is 95% sensitive (catches 95% of true cases) and 99% specific (correctly clears 99% of healthy people). If your test comes back positive, what's the actual probability you have the disease?

  • • True positive rate: 1% × 95% = 0.0095
  • • False positive rate: 99% × 1% = 0.0099
  • • P(disease | positive test) = 0.0095 ÷ (0.0095 + 0.0099) ≈ 49%

This counterintuitive result — derived from Bayes' theorem — is why US medical guidelines recommend confirmatory testing before diagnosis. A single positive result from a low-prevalence disease still means roughly a coin-flip chance of actually having it.

6. Insurance & Financial Risk Assessment

US auto insurers use large datasets to estimate annual claim probability by driver profile. If drivers aged 18–24 have a 0.06 (6%) annual probability of filing a collision claim, an insurer covering 10,000 such drivers can model their expected exposure:

  • • Expected annual claims: 10,000 × 0.06 = 600 claims per year
  • • P(at least one claim from a group of 5 such drivers) = 1 − (0.94)⁵ ≈ 26.6%
  • • P(a specific driver files 2 claims in 2 consecutive years) = 0.06 × 0.06 = 0.36%

This same logic underlies actuarial science — the statistical discipline behind every insurance premium you pay.

Understanding Your Results

What Probability Values Mean

Probability measures the likelihood that an event will occur, expressed as a number between 0 and 1:

  • 0 means impossible — the event cannot occur under any circumstances.
  • 1 means certain — the event will always occur.
  • 0.5 means an even chance — like flipping a fair coin.
  • Values between 0 and 1 represent varying levels of likelihood, often expressed as percentages by multiplying by 100.
  • Decimal vs. percentage: A probability of 0.25 equals 25%. Our calculator shows both formats automatically.

Basic Probability Formula

P(Event) = Number of Favorable Outcomes / Total Number of Outcomes

P(3) = 1/6 ≈ 0.1667 (16.67%)

Combinations (Order Doesn't Matter)

C(n, r) = n! / (r!(n - r)!)

Example:

Selecting 2 cards from a 5-card deck:

C(5, 2) = 10 possible combinations.

Conditional Probability

Conditional probability determines the likelihood of an event A occurring given that another event B has already occurred.

P(A|B) = P(A ∩ B) / P(B)

Example:

If 40% of people like coffee, and 25% like both coffee and tea:

P(Tea | Coffee) = 0.25 / 0.40 = 0.625 (62.5%)

Permutations (Order Matters)

P(n, r) = n! / (n - r)!

Example:

Arranging 3 medals among 5 athletes:

P(5, 3) = 60 possible outcomes.

Interpreting Probability Values

Probability Range
Meaning
Example
0
Impossible
Drawing a 13 from a deck of 52 cards
0.25
Unlikely
Rain on a dry summer day
0.5
Even chance
Heads or tails on a fair coin
0.75
Likely
Drawing a red card from a standard deck
1
Certain
Drawing any card from the deck

Common Real-World Applications

  • Games & Lottery – Calculate odds of winning draws or dice rolls
  • Finance & Risk – Estimate market or investment risks
  • Science & Research – Predict outcomes in experiments
  • Sports Analytics – Forecast player or team performance
  • Medical Studies – Assess likelihood of health outcomes

Mutually Exclusive, Independent & Dependent Events

Choosing the wrong event type is the most common probability mistake. Here's how to tell them apart — and which formula to use for each.

Mutually Exclusive Events

Two events are mutually exclusive if they cannot both occur in the same trial. When A happens, B is automatically ruled out. Mutually exclusive events have no overlap in the sample space.

Formula: P(A OR B) = P(A) + P(B)

Example: Rolling one die

P(rolling a 2 OR a 5) = 1/6 + 1/6 = 2/6 = 0.333 (33.3%)

You cannot roll a 2 and a 5 on the same roll — they're mutually exclusive.

Independent Events

Two events are independent if the outcome of one has no effect on the probability of the other. Repeated coin flips, separate dice rolls, and drawing with replacement are all independent.

Formula: P(A AND B) = P(A) × P(B)

Example: Coin flip AND die roll

P(Heads AND rolling a 4) = 1/2 × 1/6 = 1/12 ≈ 0.0833 (8.33%)

What the coin lands on has no influence on what the die shows.

Dependent Events

Events are dependent when the first outcome changes the probability of the second. Sampling without replacement always creates dependent events because the pool shrinks after each draw.

Formula: P(A AND B) = P(A) × P(B|A)

Example: Drawing 2 aces without replacement

P(1st Ace) = 4/52

P(2nd Ace | 1st was Ace) = 3/51

P(both Aces) = 4/52 × 3/51 = 12/2,652 ≈ 0.0045 (0.45%)

The Complement Rule

The complement of event A (written A′) is everything that is not A. Because all probabilities in a sample space sum to 1, the complement rule is often the fastest path to your answer — especially for "at least one" problems.

Formula: P(A′) = 1 − P(A)

Example: At least one head in 4 coin flips

P(no heads) = (1/2)⁴ = 1/16

P(at least one head) = 1 − 1/16 = 15/16 ≈ 0.9375 (93.75%)

Much faster than summing P(1 head) + P(2 heads) + P(3 heads) + P(4 heads).

The Addition Rule — Union Probability

For two events that are not mutually exclusive (they can both occur), you must subtract the overlap to avoid double-counting the intersection. This is the general addition rule for any two events:

P(A OR B) = P(A) + P(B) − P(A AND B)

Example: Drawing a heart OR a face card from a 52-card deck

  • • P(Heart) = 13/52 — there are 13 hearts in the deck
  • • P(Face card) = 12/52 — Jack, Queen, King in all 4 suits = 12 cards
  • • P(Heart AND Face card) = 3/52 — only 3 face cards are hearts (J♥, Q♥, K♥)
  • • P(Heart OR Face card) = 13/52 + 12/52 − 3/52 = 22/52 ≈ 0.423 (42.3%)

If you had skipped the subtraction, you'd get 25/52 ≈ 48.1% — incorrectly counting the 3 heart face cards twice.

Related Calculators

Explore these free tools to tackle related math and statistics problems.

Frequently Asked Questions

Get clear answers to the most common probability questions

For two independent events, multiply their individual probabilities: P(A AND B) = P(A) × P(B). Events are independent when the outcome of one does not affect the other — separate dice rolls, repeated coin flips, or drawing with replacement. Example: P(flipping Heads AND rolling a 6) = 1/2 × 1/6 = 1/12 ≈ 0.0833 (8.33%). This multiplication rule extends to any number of independent events chained together.

The critical difference is whether order matters. Combinations count unordered selections — {Alice, Bob} and {Bob, Alice} are the same combination, so they count once. Use combinations for teams, lottery numbers, and jury selection. Permutations count ordered arrangements — {Alice 1st, Bob 2nd} and {Bob 1st, Alice 2nd} are different outcomes, so both count. Use permutations for rankings, passwords, and race finishing positions. Formulas: C(n,r) = n! ÷ (r!(n−r)!) and P(n,r) = n! ÷ (n−r)!. P(n,r) always equals C(n,r) × r! because each unordered group has r! possible orderings.

Step 1: Identify event A (the outcome you want) and event B (the condition that is already given). Step 2: Find P(A ∩ B) — the probability both events occur together. Step 3: Find P(B) — the probability the condition itself is true. Step 4: Divide: P(A|B) = P(A ∩ B) ÷ P(B). Example using cards: P(Ace | card is a Spade) = P(Ace AND Spade) ÷ P(Spade) = (1/52) ÷ (13/52) = 1/13 ≈ 7.69%. The condition "it is a spade" narrows your effective sample space from 52 cards to just 13.

Theoretical probability is calculated using formulas and assumes perfectly fair conditions — a fair coin has exactly P(Heads) = 0.5. Experimental probability (also called empirical probability) is based on actual observed results — if you flip a real coin 1,000 times and get 487 heads, the experimental probability is 487 ÷ 1,000 = 0.487. The Law of Large Numbers guarantees that experimental probability converges toward theoretical probability as the number of trials grows. This calculator computes theoretical probability based on the values you input.

It depends on whether the events are independent or dependent. For independent events: multiply their probabilities together — P(A) × P(B) × P(C). For dependent events: use conditional probability at each step — P(A) × P(B|A) × P(C|A∩B). For "at least one" scenarios, the complement rule is fastest: P(at least one event occurs) = 1 − P(none of them occur). Example: P(at least one 6 in three die rolls) = 1 − (5/6)³ = 1 − 125/216 ≈ 0.421 (42.1%).

A probability of 0.5 means the event has an exactly equal chance of occurring or not occurring — a 50/50 split. The classic example is flipping a fair coin: heads and tails are equally likely. Probabilities always range from 0 (impossible) to 1 (certain). To convert any decimal probability to a percentage, multiply by 100: 0.5 = 50%, 0.1 = 10%, 0.75 = 75%. When someone says something is a "50-50 chance," they are expressing a probability of exactly 0.5.

Lottery odds use the combination formula because the order you match numbers does not matter. For a pick-6 lottery from numbers 1–49: C(49, 6) = 49! ÷ (6! × 43!) = 13,983,816. Your jackpot probability is 1 ÷ 13,983,816 ≈ 0.0000000715 (about 1 in 14 million). For Powerball — pick 5 from 69 white balls, plus 1 from 26 red balls: C(69, 5) × 26 = 11,238,513 × 26 = 292,201,338. Jackpot odds: roughly 1 in 292 million. Enter n = 69, r = 5 in the Combinations tab of this calculator to confirm the first part of that calculation.

Yes — probability of exactly 0 means the event is impossible, and a probability of exactly 1 means the event is certain. Example of P = 0: rolling a 7 on a standard six-sided die (there is no 7, so it cannot happen). Example of P = 1: rolling a number between 1 and 6 on that same die (one of the six faces must come up). Probability can never be negative or greater than 1. If your calculation produces a value outside [0, 1], check that your favorable outcomes do not exceed your total outcomes, and that all inputs are non-negative.

The complement rule states that P(A does NOT occur) = 1 − P(A). It works because all probabilities in a sample space must sum to 1. The rule is especially powerful for "at least one" problems: instead of summing many individual cases, calculate the single probability that none of them occur, then subtract from 1. Example: P(at least one head in 4 flips) = 1 − P(all tails) = 1 − (1/2)⁴ = 1 − 1/16 = 15/16 ≈ 0.9375 (93.75%). This approach is far simpler than computing P(exactly 1 head) + P(exactly 2 heads) + P(exactly 3 heads) + P(exactly 4 heads).