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Math & School
Probability Calculator — A and B, A or B, Not A, At Least Once
Combine the odds of independent events the way a stats class, a game designer or a risk model does
Probability is the math of chance — a number between 0 and 1 (or 0% and 100%) that says how likely something is to happen. It runs quietly under dice and card games, weather forecasts, insurance pricing, A/B tests, drug trials, loot-box drop rates and every "what are the odds" question you've ever asked. The hard part isn't a single event — it's combining two or more, and that's exactly where intuition fails most people.
This calculator works with two independent events, A and B — independent meaning one happening doesn't change the odds of the other (a coin flip doesn't remember the last flip). Enter each probability as a decimal like 0.5 or as a percent like 50%; the tool reads both. From P(A) and P(B) it computes every standard combination:
- A and B (both happen): P(A) × P(B). The classic multiplication rule for independent events.
- A or B (at least one happens): P(A) + P(B) − P(A) × P(B). You subtract the overlap so the shared chance isn't counted twice.
- Not A (A fails): 1 − P(A). The complement rule.
- Neither (both fail): (1 − P(A)) × (1 − P(B)).
- Exactly one of the two: P(A)×(1 − P(B)) + (1 − P(A))×P(B).
A quick sanity check with two fair coins, P(A) = P(B) = 0.5: A and B = 0.5 × 0.5 = 0.25 (both heads), A or B = 0.5 + 0.5 − 0.25 = 0.75 (at least one head), not A = 0.5, neither = 0.25, exactly one = 0.50. Those five add up the way they should, which is a good habit to check.
The calculator also answers the question that trips up gamers and gamblers alike: "What's the chance of at least one success if I try N times?" The trick is the complement — instead of adding up every winning case, find the chance of failing every time and subtract from one: P(at least once) = 1 − (1 − p)ⁿ. A 1-in-100 (p = 0.01) drop over 100 tries is not a guarantee — it's 1 − 0.99¹⁰⁰ ≈ 63.4%. That gap between intuition and math is why this formula is worth memorizing.
The biggest pitfall is assuming independence when events are actually linked — drawing cards without replacement, or correlated risks. If one event changes the odds of the other, the multiplication rule above doesn't apply. Use this tool for genuinely independent events; for conditional or dependent ones you'll need the full conditional-probability rules.
📖 See also: How to Calculate a Tip (and Split the Bill)
Calculator
Fill in the fields and click "Calculate" for instant results.
📰 Formula
For two independent events with probabilities P(A) and P(B): • A and B (both): P(A) × P(B) • A or B (at least one): P(A) + P(B) − P(A) × P(B) • Not A: 1 − P(A) • Neither: (1 − P(A)) × (1 − P(B)) • Exactly one: P(A)(1 − P(B)) + (1 − P(A))P(B) • At least once in n trials: 1 − (1 − p)ⁿ
📰 Formula
For two independent events with probabilities P(A) and P(B): • A and B (both): P(A) × P(B) • A or B (at least one): P(A) + P(B) − P(A) × P(B) • Not A: 1 − P(A) • Neither: (1 − P(A)) × (1 − P(B)) • Exactly one: P(A)(1 − P(B)) + (1 − P(A))P(B) • At least once in n trials: 1 − (1 − p)ⁿ
🧪 Worked examples
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Example 2
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Example 3
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Example 4
⚠️ Common mistakes
- Adding P(A) and P(B) for 'A or B' without subtracting the overlap P(A)×P(B).
- Assuming events are independent when they're not (cards drawn without replacement, correlated risks).
- Treating a 1-in-N drop over N tries as a guarantee — it's about 63%, not 100%.
- Mixing decimals and percents — enter 0.25 or 25%, not 25 as a decimal.
- Entering a probability above 1 (or above 100%), which is impossible.
💡 Tips
- Enter each probability as a decimal (0.25) or a percent (25%) — the calculator accepts both.
- For 'at least one' questions, work with the complement: 1 − (chance of failing every time).
- Check your work: not A + A and B should relate sensibly, and exactly-one + both + neither = 1.
- If one event changes the odds of the other, the events are dependent — this independent-events tool won't apply.
Embed this calculator on your site
Copy the code below and paste it into the HTML of your site or blog.
<iframe src="https://www.calcnimbus.com/embed/probability-calculator" width="100%" height="500" frameborder="0" style="border:1px solid #eee;border-radius:12px"></iframe>
❓ Frequently asked questions
How do I calculate the probability of two events both happening?
For independent events, multiply their probabilities: P(A and B) = P(A) × P(B). For example, two fair coins both landing heads is 0.5 × 0.5 = 0.25, or 25%.
How do I calculate the probability of A or B?
Use P(A or B) = P(A) + P(B) − P(A) × P(B). You add the two probabilities, then subtract the chance they both happen so the overlap isn't counted twice. With two fair coins that's 0.5 + 0.5 − 0.25 = 0.75.
What does 'independent events' mean?
Two events are independent when one happening doesn't change the probability of the other. Separate coin flips and dice rolls are independent; drawing cards without putting them back is not, because each draw changes what's left in the deck.
How do I find the chance of 'at least one' success over many tries?
Use the complement: P(at least one) = 1 − (1 − p)ⁿ, where p is the chance per try and n is the number of tries. It's far easier than adding up every winning combination, and it's exact for independent trials.
If something is 1-in-100, am I guaranteed it in 100 tries?
No. The chance is 1 − (1 − 0.01)¹⁰⁰ = 1 − 0.99¹⁰⁰ ≈ 63.4%. Roughly a one-in-three chance you'll still have nothing after 100 tries, which surprises a lot of people.
Can I enter probabilities as percentages?
Yes. Type 25% or 0.25 — both are read the same way. Just don't enter 25 on its own as a decimal, since the calculator would treat that as 2500%, which is impossible.
What's the difference between 'A or B' and 'exactly one'?
'A or B' includes the case where both happen; 'exactly one' excludes it. Exactly one = P(A)(1 − P(B)) + (1 − P(A))P(B), while A or B = that plus P(A)×P(B).
Why do the outcomes add up to 1?
For two events, the four mutually exclusive outcomes — both, only A, only B, and neither — cover every possibility, so their probabilities sum to 1. That's a quick way to check your numbers.
Does this work for dependent or conditional events?
No. This calculator uses the independent-events rules. If one event changes the odds of the other — like cards drawn without replacement — you need conditional probability, P(A and B) = P(A) × P(B given A).