Probability Calculator — Single Events, Multiple Events & Combinations
Calculate probability for single events, independent/dependent events, conditional probability, and combinations. Includes probability rules and examples.
Probability is fundamentally about counting — how many ways can something happen divided by how many total outcomes exist. The calculations themselves aren't complicated, but setting up the problem correctly is where most people go wrong. The CalcHub Probability Calculator handles the math once you've framed the question right.
What It Calculates
- Single event probability (P(A))
- Complement probability (P(not A))
- Two independent events: P(A and B), P(A or B)
- Conditional probability: P(A | B)
- Combinations C(n, r) and permutations P(n, r)
The Basics
Simple probability:P(event) = favorable outcomes ÷ total outcomes
Flip a coin: P(heads) = 1/2 = 0.5 = 50%
Roll a die, get a 4: P(4) = 1/6 ≈ 16.7%
P(not A) = 1 − P(A)
Probability of NOT rolling a 4 = 1 − 1/6 = 5/6 ≈ 83.3%
Independent vs. Dependent Events
| Event Type | Rule | Example |
|---|---|---|
| Independent (A and B) | P(A) × P(B) | Two coin flips both landing heads |
| Independent (A or B) | P(A) + P(B) − P(A and B) | Getting heads OR tails on one flip |
| Dependent | P(A) × P(B given A) | Drawing two aces without replacement |
- P(first ace) = 4/52
- P(second ace | first was ace) = 3/51
- P(both aces) = (4/52) × (3/51) = 12/2652 ≈ 0.45%
Combinations and Permutations
When order doesn't matter (like lottery numbers), use combinations:
C(n, r) = n! ÷ (r! × (n−r)!)
How many ways to choose 3 people from a group of 10?
C(10, 3) = 10! ÷ (3! × 7!) = 120
When order matters (like a PIN or race placements), use permutations:
P(n, r) = n! ÷ (n−r)!
How many ways to arrange 3 people from a group of 10 in a ranked order?
P(10, 3) = 10! ÷ 7! = 720
Conditional Probability
P(A | B) is the probability of A given that B has already occurred:
P(A | B) = P(A and B) ÷ P(B)
Medical example: A test is 95% accurate. The disease affects 1% of the population. If someone tests positive, what's the probability they actually have it?
This is Bayes' theorem territory — and the answer is surprisingly low (about 16%) because the disease is rare. Counter-intuitive results like this are exactly why working through the math matters rather than relying on gut feeling.
Probability Tables for Common Scenarios
| Scenario | Probability |
|---|---|
| Rolling at least one 6 in two die rolls | 1 − (5/6)² ≈ 30.6% |
| At least one head in 3 coin flips | 1 − (1/2)³ = 87.5% |
| Drawing a red card from a deck | 26/52 = 50% |
| Birthday collision in a group of 23 | ≈ 50.7% |
Tips
- Always check whether events are independent before multiplying probabilities. Sampling without replacement creates dependence.
- "At least one" problems are almost always easier to solve as 1 − P(none).
- Probability values always fall between 0 and 1. If you get something outside that range, there's an error in your setup.
What's the difference between a permutation and a combination?
Permutations count arrangements where order matters (ABC ≠ BAC). Combinations count selections where order doesn't matter (ABC = BAC = CAB). The permutation count is always larger than or equal to the combination count for the same n and r.
How do I calculate "at least one" probability?
Use the complement: P(at least one) = 1 − P(none). It's almost always easier. For "at least one success in 5 tries with P(success)=0.3 each": 1 − (0.7)⁵ = 1 − 0.168 = 83.2%.
Why does P(A or B) subtract P(A and B)?
Because if you just add P(A) + P(B), you count the overlap (cases where both occur) twice. Subtracting it once corrects that double-counting. This is the inclusion-exclusion principle.
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