Calculate the probability of an event from its favorable and total outcomes, and optionally combine it with a second independent event using AND / OR.
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The probability of an event is its number of favorable outcomes divided by the total number of possible outcomes. For example, rolling a 4 on a standard six-sided die has 1 favorable outcome out of 6 total, a probability of 1/6, about 16.67%.
If you enter a second event, this calculator treats events A and B as independent (one doesn't affect the other's outcome). P(A and B), the probability both happen, is P(A) × P(B). P(A or B), the probability at least one happens, is P(A) + P(B) − P(A) × P(B), which avoids double-counting the overlap where both occur.
Event A: rolling a 4 on a six-sided die, so 1 favorable outcome out of 6, P(A) = 1/6 ≈ 16.67%. Event B: drawing a heart from a standard 52-card deck, 13 favorable outcomes out of 52, P(B) = 13/52 = 25%. Treating these as independent events: P(A and B) = P(A) × P(B) = 0.1667 × 0.25 ≈ 4.17%. P(A or B) = P(A) + P(B) − P(A)×P(B) = 0.1667 + 0.25 − 0.0417 = 0.375, or 37.5%.
Using this calculator for dependent events, like drawing two cards from the same deck without replacement, since it assumes independence; dependent events need conditional probability instead, where the second probability changes based on the first outcome.
Adding P(A) + P(B) to get P(A or B) without subtracting the overlap P(A)×P(B), doing so double-counts the outcomes where both events happen and overstates the combined probability.
Expressing probability as a raw fraction of favorable-to-unfavorable outcomes (like '1 in 5 chance of failure' meaning 1/5) instead of favorable-to-total outcomes (which would be 1/6 if there are 5 unfavorable and 1 favorable, i.e. 6 total).
No, this calculator assumes events A and B are independent, meaning the outcome of one doesn't affect the other. Dependent events (like drawing cards without replacement) require conditional probability, which isn't covered by this calculator.
No, leave event B's fields at 0 to calculate the probability of event A alone.
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