Probability of 3 Events Calculator
Print PageA Probability of 3 Events Calculator (also known as a P(A, B, C) Calculator, 3-Event Inclusion-Exclusion Utility, Intersection of 3 Events Analyzer, or At-Least-One of 3 Events Odds Calculator) computes the exact probability of all three events occurring simultaneously (P(A ∩ B ∩ C)), at least one event occurring (P(A ∪ B ∪ C)), exactly one or two events occurring, or none of the events occurring across three independent, dependent, or mutually exclusive events A, B, and C.
When analyzing three events, simple addition fails because overlapping intersections create double and triple counting. Calculating the true union probability requires applying the 3-Event Inclusion-Exclusion Principle, which sums individual probabilities, subtracts pair-wise intersections, and adds back the central triple intersection.
Our free online Probability of 3 Events Calculator provides instant calculations across all standard event relationship types:
- Intersection of 3 Independent Events:
P(A ∩ B ∩ C) = P(A) · P(B) · P(C). - Chain Rule for 3 Dependent Events:
P(A ∩ B ∩ C) = P(A) · P(B | A) · P(C | A ∩ B). - 3-Event Inclusion-Exclusion Principle (Union P(A ∪ B ∪ C)):
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C). - Probability of At Least One Event (3 Independent Events):
P(At Least 1) = 1 - [ (1 - P(A)) · (1 - P(B)) · (1 - P(C)) ]. - Probability of None of the 3 Events Occurring:
P(None) = (1 - P(A)) · (1 - P(B)) · (1 - P(C)).
Master 3-Event Probability Scenario Reference Table
The table below displays the mathematical formulas, conditions, and calculated outcome percentages across common 3-event probability scenarios:
| 3-Event Calculation Target | Mathematical Formula | Event Input Conditions | Real-World Practical Benchmark | Calculated Outcome % |
|---|---|---|---|---|
| All 3 Occur (Independent) | P(A) · P(B) · P(C) |
Independent events | Basketball player makes 3 free throws (70%, 80%, 60%) | 0.70 × 0.80 × 0.60 = 33.60% |
| All 3 Occur (Dependent) | P(A) · P(B|A) · P(C|A∩B) |
Sequential sampling without replacement | Drawing 3 consecutive Aces from a 52-card deck | 4/52 × 3/51 × 2/50 = 0.0181% (1 in 5,525) |
| At Least 1 Occurs (Union) | 1 - [ (1-PA)(1-PB)(1-PC) ] |
Independent events | At least 1 free throw made (70%, 80%, 60%) | 1 – (0.30 × 0.20 × 0.40) = 97.60% |
| None of the 3 Occurs | (1-PA) · (1-PB) · (1-PC) |
Independent events | Missing all 3 free throws (70%, 80%, 60%) | 0.30 × 0.20 × 0.40 = 2.40% |
Step-by-Step 3 Free-Throw & 3-Card Drawing Calculations
To calculate the probability of a basketball player making all 3 free throws or at least 1 free throw when P(A) = 70%, P(B) = 80%, and P(C) = 60% (Independent Events), and calculate the joint probability of drawing 3 consecutive Aces from a 52-card deck without replacement (Dependent Events):
Step 1 (All 3 Independent Free Throws): P(A ∩ B ∩ C) = 0.70 × 0.80 × 0.60 = 0.3360 = 33.60%
Step 2 (None Free Throws Made): P(None) = (1 - 0.70) × (1 - 0.80) × (1 - 0.60) = 0.30 × 0.20 × 0.40 = 0.0240 = 2.40%
Step 3 (At Least 1 Free Throw Made): P(At Least 1) = 1 - P(None) = 1 - 0.0240 = 0.9760 = 97.60%
Step 4 (3 Dependent Aces - 1st Ace): P(A) = 4 ÷ 52 = 1 ÷ 13 ≈ 0.076923
Step 5 (3 Dependent Aces - 2nd Ace): P(B | A) = 3 ÷ 51 = 1 ÷ 17 ≈ 0.058824
Step 6 (3 Dependent Aces - 3rd Ace): P(C | A ∩ B) = 2 ÷ 50 = 1 ÷ 25 = 0.040000
Step 7 (Chain Rule Multiplication): P = (4/52) × (3/51) × (2/50) = 24 ÷ 132,600 = 1 ÷ 5,525 ≈ 0.0001810 = 0.0181%
Thus, the player has a 33.60% chance of making all 3 free throws (and a 97.60% chance of making at least 1), while drawing 3 Aces in a row occurs only 0.0181% of the time (1 in 5,525 attempts).
Understanding the 3-Event Inclusion-Exclusion Principle
Below is a comparative reference chart detailing why individual event probabilities cannot be directly summed when overlapping intersections exist:
| Inclusion-Exclusion Step | Mathematical Operation | Why This Step Is Required |
|---|---|---|
| Step 1: Sum Single Events | + P(A) + P(B) + P(C) |
Counts all individual event outcomes, but double counts pairs and triple counts center. |
| Step 2: Subtract Pair Intersections | - P(A∩B) - P(B∩C) - P(A∩C) |
Removes double counted pair overlaps, but accidentally removes the center triple intersection entirely! |
| Step 3: Add Back Triple Intersection | + P(A∩B∩C) |
Adds back the central triple intersection so all regions are counted exactly once. |
History & Mathematics: 1754 Abraham de Moivre to 1854 George Boole
1754 Abraham de Moivre & The Doctrine of Chances
The Inclusion-Exclusion Principle was first published by French mathematician Abraham de Moivre in his 1754 third edition of The Doctrine of Chances, solving multi-event card matching probability problems.
1854 George Boole & Boolean Algebra
In 1854, English mathematician George Boole published An Investigation of the Laws of Thought, establishing symbolic Boolean algebra for set unions (A ∪ B ∪ C) and intersections (A ∩ B ∩ C).
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Frequently Asked Questions (FAQ)
How do you calculate the probability of 3 independent events happening together?
Multiply their probabilities together: P(A ∩ B ∩ C) = P(A) × P(B) × P(C).
What is the Inclusion-Exclusion formula for 3 events?
The formula is P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(A ∩ C) + P(A ∩ B ∩ C).
How do you calculate the probability of AT LEAST ONE of 3 independent events occurring?
Use the complement rule: P(At Least 1) = 1 - [ (1 - P(A)) × (1 - P(B)) × (1 - P(C)) ].