🎲 Probability Calculator
Calculate any probability - single events, AND/OR combined events, conditional probability, "at least one" outcomes, binomial trials, combinations, permutations, dice, and coin flips - with complete step-by-step solutions for every calculation. Enter your scenario and see both the formula and the working.
🎲 Select Calculation Type
🎲 Dice Roll Simulator
🪙 Coin Flip Calculator
🃏 Card Draw Probability
🔢 Combinations & Permutations
🔡 Permutations with Repetition
📊 Factorial Calculator
📐 Probability Formulas
Basic Probability
Range: 0 ≤ P(A) ≤ 1
P(impossible) = 0
P(certain) = 1
P(A) + P(not A) = 1
AND (Intersection)
Dependent events: P(A∩B) = P(A) × P(B|A)
Example: P(head AND 6) = 0.5 × (1/6) = 1/12
OR (Union)
Mutually exclusive: P(A∪B) = P(A) + P(B)
Example: P(Heart OR Face card)
= 13/52 + 12/52 − 3/52 = 22/52 ≈ 42.3%
Conditional Probability
Read as: "Probability of A given B has occurred"
Bayes' Theorem:
P(A|B) = P(B|A) × P(A) / P(B)
Combinations & Permutations
C(n,r) = n! / (r! × (n−r)!)
Permutation (order matters):
P(n,r) = n! / (n−r)!
With repetition: nʳ arrangements
Binomial Distribution
Where: n = trials, k = successes, p = prob success
Mean = n×p
Std Dev = √(n×p×(1-p))
Example: P(exactly 3 heads in 10 flips)
= C(10,3) × 0.5³ × 0.5⁷
= 120 × 0.125 × 0.0078125 = 11.72%
❓ Frequently Asked Questions
Probability Calculator - Core Rules, Formulas and Real-World Applications
Probability is the mathematical language of uncertainty. From rolling dice to medical testing, from quality control to financial modelling, probability underlies every area where decisions are made with incomplete information. The formulas are relatively simple - but applying the right formula to the right scenario requires understanding the distinctions between event types, independence, and when to use AND vs OR rules.
AND vs OR - The Two Most Important Probability Rules
AND - Both Events Occur (Intersection)
- Independent events: P(A and B) = P(A) × P(B)
- Coin flip then die roll: P(heads AND 6) = 0.5 × 1/6 = 1/12 ≈ 8.3%
- Dependent events: P(A and B) = P(A) × P(B|A)
- Drawing 2 aces from a deck without replacement: P(A₁) × P(A₂|A₁) = 4/52 × 3/51 = 0.452%
- Each successive event changes the sample space
OR - At Least One Event Occurs (Union)
- Formula: P(A or B) = P(A) + P(B) − P(A and B)
- Subtract overlap to avoid double-counting
- P(red card OR face card) = 26/52 + 12/52 − 6/52 = 32/52 = 61.5%
- Mutually exclusive events: P(A or B) = P(A) + P(B) - no overlap to subtract
- P(rolling 1 OR 6) = 1/6 + 1/6 = 2/6 = 33.3% (can't roll both at once)
The "At Least One" Shortcut - Using the Complement
Calculating "at least one" directly is complex - you'd need to add P(exactly 1) + P(exactly 2) + ... + P(exactly n). The complement approach is far simpler: P(at least one) = 1 − P(none).
Example: P(at least one head in 5 coin flips). Direct method: sum 5 terms. Complement method: P(no heads) = (0.5)⁵ = 0.03125. P(at least one head) = 1 − 0.03125 = 96.875%. Same answer, one calculation. This works because "at least one" and "none" are complementary events - their probabilities must add to 1.
Combinations vs Permutations - The Key Distinction
Both count ways to select items from a group. The distinction is whether order matters:
- Combinations C(n,r) = n! ÷ (r! × (n-r)!) - Order doesn't matter. Choosing a committee of 3 from 10 people: C(10,3) = 120. The committee {A,B,C} is the same as {B,A,C}.
- Permutations P(n,r) = n! ÷ (n-r)! - Order matters. Arranging 3 people in 3 specific positions from 10: P(10,3) = 720. President/VP/Secretary: {A,B,C} is different from {B,A,C}.
- Relationship: P(n,r) = C(n,r) × r! - the number of permutations equals the number of combinations multiplied by the number of ways to arrange r items.
Binomial Probability - Repeated Independent Trials
When you repeat an experiment n times where each trial has the same probability p of success (and q = 1−p of failure), the probability of exactly k successes follows the binomial distribution:
P(exactly k successes in n trials) = C(n,k) × p^k × (1−p)^(n−k)
Example: P(exactly 3 heads in 10 fair coin flips). C(10,3) = 120. p^3 = 0.5³ = 0.125. (1−p)^7 = 0.5⁷ = 0.0078125. P = 120 × 0.125 × 0.0078125 = 11.72%. The binomial distribution applies whenever: trials are independent, each trial has the same two outcomes, and p is constant. Coin flips, quality control testing, medical trial outcomes, and survey responses all follow this model.