How it works

A useful distinction

At least one avoids double counting

With three independent trials each having a 10% success chance, the chance of at least one success is 1 − 0.9³ = 27.1%. Adding 10% three times counts some multiple-success outcomes more than once.

Checking dependence and event definitions

For independent events with probabilities 0.4 and 0.3, the chance of both is 0.4 × 0.3 = 0.12. The chance of at least one is 0.4 + 0.3 − 0.12 = 0.58, because outcomes where both happen must not be counted twice. Independence is an assumption, not something established by entering two probabilities. If the events are dependent, use their actual intersection or conditional information rather than the independent-event shortcut. Probability fields use decimals from 0 to 1.

Two events: formula and assumptions

P(A or B) = P(A) + P(B) − P(A and B). Independent events have P(A and B) = P(A)P(B).

For dependent events, enter the intersection. It must satisfy probability bounds.

Normal interval: formula and assumptions

P(a ≤ X ≤ b) = Φ((b−μ)/σ) − Φ((a−μ)/σ).

This model assumes a continuous normal distribution. The results also show central coverage intervals, which are not confidence intervals for an estimated mean.

Normal probability to boundary: formula and assumptions

Boundary=mean+standard deviation×inverse normal probability.

Select the tail whose probability is given.

Solve independent-event probabilities: formula and assumptions

For independent events p=P(A), q=P(B): intersection=pq, union=p+q−pq, exactly one=p+q−2pq, neither=(1−p)(1−q).

Enter two different constraints. Aggregate constraints can have two labeled solutions; redundant constraints can leave infinitely many possibilities. Independence is required.

Repeated independent trials: formula and assumptions

All successes=p^n; no successes=(1−p)^n; at least one=1−(1−p)^n. Joint independent groups multiply.

Each trial and the two groups are independent, with a fixed probability within each group. No simulated trials are used.

Worked example

Using the Two events example:

Inputs

P(A): 0.4
P(B): 0.3
Relationship: Independent

Method

P(A or B) = P(A) + P(B) − P(A and B). Independent events have P(A and B) = P(A)P(B).

Result

Example result

P(A or B): 0.58

P(A and B): 0.12

P(neither): 0.42

P(exactly one): 0.46

These are illustrative inputs. Change the values to explore a different scenario.

Formula references

References explain the underlying methods; they do not imply endorsement or individual professional review of Finflo.

Common questions

When can I multiply two event probabilities?

Multiplication gives the intersection directly when the events are independent. Dependent events require appropriate joint or conditional information.

Why can the independent-event solver return two solutions?

Some aggregate constraints fit two labeled event pairs. Redundant constraints can also leave infinitely many possibilities rather than a unique pair.

Prepared by Finflo. See our methodology and editorial policy.

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