155
Two events, four answers
On the default — a 60% chance of rain and a 40% chance of traffic: both happen 24% of the time. Enter your own two events below.
FREE TOOL · STATISTICS
Probability of Two Events
Enter the chances of two events and get all four answers at once: both happen, either happens, neither happens, and exactly one happens. Switch to conditional mode when the second event depends on the first — cards without replacement, weather chains, medical tests.
The four questions, one page
For two independent events — rain and traffic, a coin and a die — one does not change the other, so the multiplication rule does all the work: P(A and B) = P(A) × P(B). On the defaults (60% and 40%), that is 0.60 × 0.40 = 24%. The other answers follow from the same two numbers: either means add, then subtract the overlap so the both-cases are not counted twice (60 + 40 − 24 = 76%); neither is the complement — 1 minus the either-chance, or (1 − 0.60)(1 − 0.40) = 24%; exactly one is A-without-B plus B-without-A = 36% + 16% = 52%. Notice the four answers sum to 100% — they cover every possible outcome.
Output · formula · meaning
| Output | Formula | What it means |
|---|---|---|
| Both A and B | P(A) × P(B) | 0.60 × 0.40 = 24% — the intersection, the overlap. |
| Either A or B | P(A) + P(B) − P(both) | 60% + 40% − 24% = 76% — the union; the subtraction keeps the overlap counted once. |
| Neither | (1 − P(A)) × (1 − P(B)) | 0.40 × 0.60 = 24% — the complement of either. |
| Exactly one | P(A)×(1−P(B)) + P(B)×(1−P(A)) | 36% + 16% = 52% — XOR, the exclusive either. |
Worked example: a coin and a die
P(heads) = 50%, P(six) = 16.67%. Both: 0.50 × 0.1667 = 8.33% — about once in twelve throws. Either: 50 + 16.67 − 8.33 = 58.33%. Neither: 50% × 83.33% = 41.67%. Exactly one: 41.67% + 8.33% = 50% — a coin-toss-worthy coincidence of the math, not a rule.
Independent vs dependent: when to use conditional mode
Independence means the first event does not change the second — coins, dice, and weather on different days. Drawing two cards without replacement breaks that: the first draw changes the deck. Then you need the conditional probability P(B|A), and the rule becomes P(A and B) = P(A) × P(B|A). Two aces in a row: P(first ace) = 4/52 = 7.69%; P(second ace given the first) = 3/51 = 5.88%; both = 7.69% × 5.88% = 0.45%. Switch the calculator to dependent mode and enter the 5.88 yourself — it multiplies correctly where the naive 7.69% × 7.69% = 0.59% would overstate it.
The “at least one” shortcut
“What is the chance that at least one happens?” sounds hard and is easy: it is just 1 minus the chance that neither happens. On the defaults, 1 − 24% = 76% — the same as the “either” answer, because “at least one” and “either” describe the same outcomes. Whenever a question says “at least one,” compute the neither-case first and subtract from 100%.
Who uses this
- Students: “both / either / neither” probability questions are exam staples — with the steps shown for partial credit.
- Game designers & players: two-drop odds, crit chains, and card-draw probabilities without the spreadsheet.
- Analysts: combining failure rates, conversion funnels, and A/B test overlap questions.
- Teachers: demonstrating independence vs dependence with the card-without-replacement example.
Accuracy & limitations
All results use the exact decimals you enter — no floating-point shortcuts — and percents display to two decimal places. Independent mode assumes the events do not influence each other; if they do, dependent mode is required and you must supply P(B|A). Inputs must be percents between 0 and 100. For three events, see the probability of three events calculator.
Rolling dice to feed this calculator? The dice roller gives the odds and expected value of any throw, and the fraction to percent converter turns odds like 1/6 into the percents this page expects.
Cite this calculatorBright Side Kit. “Probability of Two Events Calculator.” brightoolbox.com, 2026.
Bright Side Kit. "Probability of Two Events Calculator." https://brightoolbox.com/probability-two-events/ Accessed October 22, 2026.Frequently asked questions
How do you find the probability of two independent events both happening?
Multiply them. P(A and B) = P(A) × P(B). A 60% chance of rain and a 40% chance of traffic mean a 0.60 × 0.40 = 24% chance of both.
How do you find the probability of A or B?
Add, then subtract the overlap: P(A or B) = P(A) + P(B) − P(A and B). Skipping the subtraction double-counts the cases where both happen. On the defaults: 60 + 40 − 24 = 76%.
What is the probability of at least one of two events?
One minus the neither-chance: 1 − P(neither). At least one, either, and the union are all the same question — 76% on the defaults.
What is the difference between independent and dependent events?
Independent events do not affect each other (coin flips, dice) — multiply directly. Dependent events do (cards without replacement, tests after symptoms) — you need the conditional probability P(B|A), and P(A and B) = P(A) × P(B|A).
How do you find conditional probability P(B|A)?
From data: P(B|A) = P(A and B) ÷ P(A). Or flip it: if you know P(B|A), then P(A and B) = P(A) × P(B|A) — the dependent mode of the calculator above.
What is the probability of two heads in two coin flips?
25%. 0.50 × 0.50 = 0.25 — the textbook independent-events case.
What is the probability of exactly one of two events?
Add the two one-sided cases: P(A)×(1−P(B)) + P(B)×(1−P(A)). On the defaults: 36% + 16% = 52%.