What Is Probability of 3 Events?
When three events are independent, the probability that they all occur is the product of the three individual probabilities. The probability that none occur is the product of the three complements, and "at least one" is one minus that. The calculator applies these rules for you and lists the result for every count from zero to three.
The calculator assumes the events do not influence each other. The page also shows the general inclusion-exclusion formula for the probability that any of the events occur, which is the version to use when you are not assuming independence and you have the overlap probabilities.
Scroll down for the formulas, a variables table, a worked example with three different probabilities and answers to common questions such as P(A and B and C) versus P(A or B or C).
How to Use the Probability of 3 Events Calculator
- Enter P(A), P(B) and P(C) as decimals, percentages or fractions.
- Read the headline results: all, none, and at least one.
- Use the table for exactly 0, 1, 2 or 3 events and the cumulative “at least k” column.
- If your events are not independent, use conditional probabilities instead; see the notes below.
Probability of 3 Events Formula
All three occur (independent)
None occur
At least one occurs
Inclusion-exclusion (any events)
Probability of 3 Events Variables Explained
| Symbol | Meaning | Notes |
|---|---|---|
| P(A), P(B), P(C) | Probability of each event | Between 0 and 1 (or 0 % to 100 %) |
| P(all) | All three occur | P(A) × P(B) × P(C) for independent events |
| P(none) | None occur | (1 − P(A))(1 − P(B))(1 − P(C)) |
| P(≥1) | At least one occurs | 1 − P(none) |
Probability of 3 Events Worked Example
With the example values loaded in the calculator above, the steps are:
- Assuming independence, P(all) = 0.5 × 0.3 × 0.2 = 0.03.
- P(none) = (1 − 0.5) × (1 − 0.3) × (1 − 0.2) = 0.28.
- P(at least one) = 1 − P(none) = 0.72.
- P(exactly k) is found by adding the probabilities of every combination with k events happening and the rest failing: k=0: 0.28, k=1: 0.47, k=2: 0.22, k=3: 0.03.
Result: All 3 events occur P(A ∩ B ∩ C) = 0.03; None occur = 0.28; At least one occurs P(A ∪ B ∪ …) = 0.72.
Reading the Exactly-k Table
For P(A) = 0.5, P(B) = 0.3 and P(C) = 0.2, the chance that exactly one event occurs is 0.28 + 0.12 + 0.07 = 0.47, the three terms being A only, B only and C only. Exactly two is 0.12 + 0.07 + 0.03 = 0.22. With none at 0.28 and all three at 0.03, the four outcomes add to exactly 1.
The expected number of events is the sum of the three probabilities, 1.0. The most likely single outcome is exactly one event, at 0.47, although the chance of at least one event is the larger 0.72. Always check which of these questions you are really asking.
Worked example
Suppose three machines each fail on a given day with probabilities 0.5, 0.3 and 0.2. The chance all three fail is 0.5 × 0.3 × 0.2 = 0.03. The chance none fail is 0.5 × 0.7 × 0.8 = 0.28. So the chance at least one fails is 1 − 0.28 = 0.72. Exactly one failing is the sum of three cases: only A fails (0.5 × 0.7 × 0.8 = 0.28), only B fails (0.5 × 0.3 × 0.8 = 0.12) and only C fails (0.5 × 0.7 × 0.2 = 0.07), which total 0.47.
Independent, mutually exclusive and dependent events
This calculator assumes independence. Rolling dice, separate coin flips or unrelated machines are independent. Drawing cards without replacement or events caused by a common factor, such as weather, are not.
| Relationship | Meaning | P(A and B) | P(A or B) |
|---|---|---|---|
| Independent | One does not affect the other | P(A)·P(B) | P(A) + P(B) − P(A)P(B) |
| Mutually exclusive | Cannot both happen | 0 | P(A) + P(B) |
| Dependent | One changes the other’s chance | P(A)·P(B | A) | P(A) + P(B) − P(A and B) |
Why “at least one” uses the complement
Counting every way at least one event could happen is tedious. It is far easier to compute the single case where none happen and subtract from 1. This complement trick works for any number of events and is behind many “at least one” problems.
Where Probability of 3 Events Is Used
Three independent chances
Find the chance that all three, none or at least one of three events occurs.
Exactly k
See the chance of exactly 1 or exactly 2 events.
Reliability
Combine three independent failure or success probabilities.
Homework
Check each answer against the distribution table.
Common Probability of 3 Events Mistakes to Avoid
- Check independence before multiplying probabilities.
- Probabilities of different outcomes of the same single event cannot add to more than 1.
- Use the complement for “at least” problems.
Probability of 3 Events Terms and Definitions
- Independent events
- Events where the occurrence of one does not change the probability of the other.
- Complement
- The event that something does not happen; probability 1 − P.
Probability of 3 Events FAQs
How do you find the probability of three events happening?
For independent events, multiply the three probabilities: P(A) × P(B) × P(C).
How do you find the probability of at least one of three events?
Subtract the probability that none occur from 1: 1 − (1 − P(A))(1 − P(B))(1 − P(C)).
What if the events are not independent?
Use conditional probabilities, such as P(A) × P(B | A) × P(C | A and B), or the inclusion-exclusion formula with known joint probabilities.
What does exactly one mean?
Exactly one event happens and the other two do not. Add the probabilities of each such combination.
Can I enter percentages or fractions?
Yes. Type 25%, 0.25 or 1/4.
Why do the probabilities of exactly 0, 1, 2 and 3 add to 1?
They cover every possible outcome, so together they must total 100%.
Can I use this calculator for dependent events?
No. It assumes the three events are independent, so it multiplies their probabilities. For dependent events you need conditional probabilities, for example P(A and B) = P(A) × P(B given A).
How do I find the probability that A or B or C happens?
For independent events, P(at least one) = 1 − (1 − P(A))(1 − P(B))(1 − P(C)). The general inclusion-exclusion formula is P(A) + P(B) + P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C).
How do I find the chance of at least two of three events?
Add the probability of exactly two to the probability of all three. For three events that each have probability 0.5, that is 0.375 + 0.125 = 0.5. The table in the results gives the cumulative “at least k” values directly.
What is the chance of getting at least one head in three coin tosses?
Each toss has a probability of 0.5, so the chance of no heads is 0.5 × 0.5 × 0.5 = 0.125 and the chance of at least one head is 1 − 0.125 = 0.875, or 87.5%.
How do I find A and B but not C?
For independent events, multiply P(A) × P(B) × (1 − P(C)). The “exactly two” result is the sum of three such terms, one for each pair.