Probability Calculator

What this calculator answers

It answers two kinds of probability question. With two events, it finds the chance that A and B both happen, that at least one happens, that exactly one or neither happens, and the chance of one event once the other has happened. The events can be independent, mutually exclusive, or dependent, with how they overlap given as P(A and B) or P(B given A). With repeated tries, it finds the chance of at least one success in n tries at the same chance each, and how many tries a given chance takes.

How to use it

• Problem type: two events, or repeated tries.
• Probabilities as: how plain numbers are read and results shown. With decimals, 0.4 means 0.4. With percent, 40 means 40%. Fractions such as 2/5, entries with a % sign such as 40%, and 1 in 5 mean the same either way. Switching converts the numbers already typed, so 0.4 becomes 40.
• How A and B are related: independent (the default, and an assumption you should check), mutually exclusive (they can’t both happen), or dependent, where you enter either Probability of A and B or Probability of B given A. The line under the choice explains each one.
• Probability of A and Probability of B: each between 0 and 1 (0% and 100%).
• Find: which probability is the headline. Every other one is in the table below it.
• Chance per try, Number of tries (a whole number up to 1,000,000,000) and, optionally, a Target chance for repeated tries.

Results update as you type. The Try buttons load the cases worked through on this page: two sixes, two aces, a 1 or a 6 on one die, at least one six in 4 rolls, and a 1% chance over 100 tries. To compare two situations, press Save for comparison, change the inputs and save again. The comparison lists the main probabilities in percent, with each change in percentage points. For exactly k, at most k or at least k successes over repeated tries, Continue in the Binomial Distribution Calculator carries the chance per try and the number of tries over.

Probability of A and B (independent and dependent events)

Multiply the chance of A by the chance of B once A has happened:

For independent events, A happening doesn’t change the chance of B, so $P(B \mid A) = P(B)$ and the rule becomes $P(A) \times P(B)$. For dependent events, use the changed chance. Drawing two aces from a shuffled 52-card deck without putting the first card back:

1. First card an ace: P(A) = 4/52 = 1/13.
2. Second card an ace once the first was: 3 aces are left among 51 cards, so P(B given A) = 3/51 = 1/17.
3. Both aces: 1/13 × 1/17 = 1/221, about 0.00452 (0.452%).

Returning the first card and reshuffling would make the draws independent: 1/13 × 1/13 = 1/169, about 0.00592.

Probability of A or B

Add the two probabilities, then subtract the overlap so that outcomes where both happen are not counted twice:

“Or” includes the outcomes where both happen. For mutually exclusive events the overlap is 0, so the probabilities simply add: rolling a 1 or a 6 with one die is 1/6 + 1/6 = 1/3, about 0.3333.

Conditional probability: P(A given B)

Divide the chance that both happen by the chance of the event you are given:

This works only when P(B) is above 0. If B never happens, “the chance of A once B has happened” has no value, and the calculator shows “Not defined” rather than dividing by 0. P(A given B) and P(B given A) are usually different numbers, as the example below shows.

Worked example: rain on Saturday or Sunday

A forecast gives a 40% chance of rain on Saturday (A) and a 30% chance on Sunday (B). Suppose it also says that if Saturday is wet, the chance of rain on Sunday rises to 45% (B given A). What is the chance of rain on at least one day of the weekend?

1. Rain on both days: P(A and B) = 0.4 × 0.45 = 0.18, or 18%.
2. Rain on at least one day: P(A or B) = 0.4 + 0.3 − 0.18 = 0.52, or 52%.
3. No rain at all: P(neither) = 1 − 0.52 = 0.48, or 48%.
4. Rain on Saturday, given a wet Sunday: P(A given B) = 0.18 ÷ 0.3 = 0.6, or 60%.

Two shortcuts give wrong answers here. Adding 40% + 30% = 70% counts the weekends with rain on both days twice. Treating the days as independent gives 0.4 × 0.3 = 12% for both days and 40% + 30% − 12% = 58% for at least one. That is 6 percentage points too high, because a wet Saturday makes a wet Sunday more likely. The calculator flags this: with these numbers, A raises the chance of B from 30% to 45%, so the events are dependent.

Independent vs. mutually exclusive events

Independent events don’t affect each other’s chances. Mutually exclusive events can’t happen together. They are different ideas. If A and B each have a chance above 0 and can’t both happen, then A happening makes B impossible, so mutually exclusive events are dependent.

When you enter P(A and B) or P(B given A), the calculator checks whether your numbers happen to be independent, meaning P(A and B) equals P(A) × P(B), and says so either way.

Probability of at least one success in n tries

Work out the chance that every try fails, then take it away from 1. With a chance $p$ of success on each of $n$ independent tries:

The chance of at least one six in 4 rolls of a die is 1 − (5/6)⁴ = 1 − 625/1296 = 671/1296, about 0.5177 (51.77%). The chance of no six at all is 625/1296, about 0.4823. With fractions typed in, the calculator keeps the answer exact.

How many tries for a given chance?

Solve $1 - (1 - p)^n \ge t$ for the number of tries $n$, where $t$ is the chance you want, and round up to a whole try:

With a 1% chance per try:

When the ratio is exactly a whole number, computer arithmetic can land a hair above it. With an 8% chance per try and a target of 15.36%, the answer is exactly 2 tries, because 1 − 0.92² = 0.1536. But ln(1 − 0.1536) ÷ ln(1 − 0.08) comes out as 2.0000000000000004 in standard double-precision arithmetic, which rounds up to 3. The calculator checks the answer and the try before it with exact fractions, so it reports 2.

Why some inputs are impossible

Some combinations can’t describe any real situation, so the calculator names the input and says what would work instead of printing a negative probability:

• P(A and B) can’t be more than P(A) or P(B). Both happen only when each one happens.
• P(A and B) must be at least P(A) + P(B) − 1. Otherwise P(A or B) would be more than 1. With P(A) = 0.7 and P(B) = 0.6, the events must overlap by at least 0.3.
• Mutually exclusive events can’t add up to more than 1. With P(A) = 0.7, P(B) can be at most 0.3.
• P(B given A) can be at most P(B) ÷ P(A). With P(A) = 0.4 and P(B) = 0.3, that is 0.75, because P(A) × P(B given A) can’t be more than P(B).
• A probability above 1 is refused. If you type 40 while decimals are chosen, the message suggests 40% or switching to percent.

Reading the result

• The headline is the probability chosen under Find, followed by the other forms: percent or decimal, “1 in N” and, when you typed fractions, the exact fraction.
• The three tiles show three more of P(A and B), P(A or B), P(A given B) and P(B given A), leaving out the one in the headline.
• The dependence note appears when you entered P(A and B) or P(B given A), or chose mutually exclusive events. It says whether the events are independent, and what assuming independence would have given.
• The Venn diagram puts each probability in its region (A only, both, B only, neither) and shades the headline’s region. It isn’t drawn to scale.
• Every probability for A and B lists all twelve with the formula used. The 2 × 2 probability table crosses A and not A with B and not B; its totals are the single-event chances. Both download as CSV files.
• For repeated tries, the headline is the chance of at least one success. The tiles add no successes, exactly one, every try succeeding, the average number of successes and the tries for a 50% chance and for 90% (or your target). The chart shows how the chance grows with more tries. Tries needed for a given chance gives the chance after that many tries and with one try fewer, so you can see the target being crossed.

Assumptions and limitations

• Independence is only used when you choose it. The calculator never applies it silently to events you describe as dependent.
• Repeated tries assume every try is independent with the same chance. Pity timers in games (a sure drop after a run of misses) and drawing without replacement change the chance from try to try, so the formula doesn’t fit them.
• It handles two events at a time. Three or more events need more information about how they overlap.
• The answer is only as good as the probabilities you enter. A forecast’s 40% is itself an estimate.
• Fractions are carried exactly and results are rounded for display only. Very small probabilities are written in scientific notation (1E-200) rather than rounded to 0.

Common mistakes

• Adding probabilities of events that can overlap. 40% + 30% = 70% for rain on the weekend double-counts the weekends with rain on both days. The true figure in the example is 52%.
• Multiplying dependent events as if they were independent. 0.4 × 0.3 = 12% for rain on both days ignores that the days affect each other; with the stated 45%, it is 18%.
• Multiplying the chance per try by the number of tries. 1% × 100 tries is not 100%; the chance of at least one success is 63.4%. At 200 tries the shortcut would give 200%, which is impossible.
• Swapping the condition. In the example, P(A given B) is 60% but P(B given A) is 45%. Check which event you are told has happened.
• Calling mutually exclusive events independent. Events that can’t happen together are dependent, unless one of them has a chance of 0.
• Typing a percentage where a decimal is expected. 40 is not a probability between 0 and 1. Type 40% or switch to percent; the calculator catches this for you.