🎲Probability Calculator
Calculate probabilities for two independent events (union, intersection, conditional), binomial series (exact k successes, at least k, at most k), or normal distribution range probabilities. Covers all core probability scenarios in one tool.
Prefer to skip the form? Scroll down and Ask AI Instead. Just describe your situation and let AI handle the math for you in seconds.
Primary Result
1
Probability Breakdown
✦ Ask AI Instead
Probability Calculator: Two Events, Binomial Series, and Normal Distribution
Probability measures the likelihood of an event occurring on a scale from 0 (impossible) to 1 (certain). For two independent events A and B: the intersection (both occur) uses P(A ∩ B) = P(A) × P(B); the union (at least one occurs) uses P(A ∪ B) = P(A) + P(B) − P(A ∩ B). For a series of n independent trials, use the binomial distribution.
Key formulas: P(A∩B) = P(A)×P(B) | P(A∪B) = P(A)+P(B)−P(A∩B) | P(X=k) = C(n,k)×p^k×(1−p)^(n−k)
| P(A) | P(B) | P(A∩B) | P(A∪B) |
|---|---|---|---|
| 0.5 | 0.4 | 0.20 | 0.70 |
| 0.3 | 0.6 | 0.18 | 0.72 |
| 0.25 | 0.25 | 0.0625 | 0.4375 |
This calculator covers three core probability scenarios: two independent events (union, intersection, conditional, exclusive or, neither), binomial series (exact, at most, at least k successes out of n trials), and normal distribution range probabilities.
Two Independent Events
Two events are independent if the occurrence of one does not affect the probability of the other: P(A|B) = P(A). Coin flips are independent; drawing cards without replacement is not. For independent A and B:
P(A ∩ B) = P(A) × P(B) — both occur simultaneously.
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) — at least one occurs. The subtraction avoids counting the overlap twice.
P(A ⊕ B) = P(A ∪ B) − P(A ∩ B) — exactly one occurs (exclusive or). P(neither) = P(A')×P(B') = (1−P(A))×(1−P(B)). These four values always sum to 1.
Binomial Distribution
The binomial distribution models the number of successes X in n independent trials, each with success probability p. The probability of exactly k successes is P(X=k) = C(n,k) × p^k × (1−p)^(n−k), where C(n,k) = n!/(k!(n−k)!) is the binomial coefficient counting the number of ways to arrange k successes among n trials.
Example: probability of exactly 3 heads in 10 coin flips (p=0.5): P(X=3) = C(10,3) × 0.5³ × 0.5⁷ = 120 × 0.125 × 0.0078125 ≈ 0.1172. The expected number of heads is E[X] = np = 10 × 0.5 = 5. Cumulative: P(X ≤ 3) = sum of P(X=0) through P(X=3) ≈ 0.1719.
Normal Distribution Probability
For continuous outcomes following a normal distribution N(μ, σ²), the probability of a value falling between bounds L and R is P(L < x < R) = Φ(Z_R) − Φ(Z_L), where Z = (x − μ)/σ and Φ is the standard normal CDF. Unlike discrete probabilities, P(x = exactly some value) = 0 for a continuous distribution — only range probabilities have nonzero values. Use the normal distribution mode for scores, measurements, and any approximately normally distributed quantity.
Frequently Asked Questions
What is the probability that A or B occurs (union)?
For two independent events: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The subtraction prevents double-counting: if P(A)=0.5 and P(B)=0.4, simply adding gives 0.9 but this counts the overlap (both occurring) twice. P(A ∩ B) = 0.5 × 0.4 = 0.20 (for independent events), so P(A ∪ B) = 0.5 + 0.4 − 0.20 = 0.70. For mutually exclusive events (cannot both occur), P(A ∩ B) = 0 so P(A ∪ B) = P(A) + P(B) directly.
What is the binomial distribution used for?
The binomial distribution models the number of successes in n independent yes/no trials with constant success probability p. It applies when: (1) fixed number of trials n, (2) each trial has only two outcomes (success/failure), (3) trials are independent, (4) probability p is constant. Examples: number of heads in n coin flips (p=0.5); number of defective items in n manufactured parts (p=defect rate); number of patients responding to treatment in a clinical trial. The formula P(X=k) = C(n,k)p^k(1−p)^(n−k) gives the probability of exactly k successes.
How do I calculate the probability of at least one success?
P(at least one success) = 1 − P(no successes) = 1 − (1−p)^n. This is almost always easier to compute than summing P(X=1) + P(X=2) + ... + P(X=n). Example: probability of at least one 6 in four dice rolls (p=1/6, n=4): P(X≥1) = 1 − (5/6)^4 = 1 − 625/1296 ≈ 0.5177. About 51.8% chance. The probability of at least one success grows quickly with n: for p=0.1 and n=10, P(X≥1) = 1 − 0.9^10 ≈ 0.6513; for n=20 it's ≈ 0.8784.
What is the difference between P(A and B) and P(A given B)?
P(A ∩ B) is the probability that both A and B occur together. P(A | B) is the conditional probability — the probability that A occurs given that B has already occurred. They are related by: P(A|B) = P(A ∩ B) / P(B). For independent events: P(A|B) = P(A) because knowing B happened tells you nothing about A. For dependent events they differ — e.g., drawing two aces from a deck: P(2nd ace | 1st was ace) = 3/51 ≈ 0.059, but P(2nd ace) without conditioning = 4/52 ≈ 0.077. This calculator assumes independence: P(A∩B) = P(A)×P(B).
What does "independent events" mean in probability?
Two events A and B are independent if the occurrence of one provides no information about the other: P(A|B) = P(A) and P(B|A) = P(B). Equivalently, P(A ∩ B) = P(A) × P(B). Examples of independent events: two separate coin flips; rolling two dice; two customers independently deciding to buy a product. Examples of dependent events: drawing cards without replacement; weather on consecutive days; a student's performance on related exam questions. This calculator assumes independence — for dependent events, you need P(A∩B) as a separate input.