Mutual Independence: Definition, Formula and Pairwise Counterexample

Events are mutually independent when the probability of every possible combination of them occurring together equals the product of their individual probabilities. For two events that is one condition. For three or more events it is several, and checking only pairs is not enough. This guide gives the definition, the number of conditions to check, a worked counterexample with the arithmetic shown, and short code to test it.

Definition of mutual independence

A collection of events is mutually independent if, for every finite subset of the collection, the probability that all events in that subset occur equals the product of their individual probabilities (Siegrist, n.d.). It is stronger than pairwise independence, where only each pair of events must satisfy the product rule.

For two events, the familiar test applies: A and B are independent if P(A | B) = P(A), or P(B | A) = P(B), or P(A and B) = P(A)P(B) (Illowsky and Dean, n.d.). Mutual independence extends the product form to every sub-collection. For the two-event case alone, see how to calculate conditional probability.

Formula and what each symbol means

For events A₁, A₂, …, A_n to be mutually independent, the following must hold for every subset J of the indices with at least two members:

P(∩_{j in J} A_j) = ∏_{j in J} P(A_j)

• A_j: one of the events, a subset of the sample space.
• J: any chosen sub-collection of the events.
• ∩: the intersection, meaning all the chosen events occur together.
• ∏: the product of the individual probabilities.

Independence of three events

For events A, B and C, mutual independence requires four equations, not one:

1. P(A and B) = P(A)P(B)
2. P(A and C) = P(A)P(C)
3. P(B and C) = P(B)P(C)
4. P(A and B and C) = P(A)P(B)P(C)

The first three are the pairwise conditions; the fourth is the triple condition. In general, n events have 2ⁿ − n − 1 subsets of size two or more, so that is the number of product equations to check (Siegrist, n.d.). That gives 4 for n = 3, 11 for n = 4 and 26 for n = 5.

Worked example: pairwise independent but not mutually independent

Pairwise vs mutual independence

Roll two fair dice, so there are 36 equally likely outcomes. Define A: the first die shows 3; B: the second die shows 4; C: the sum is 7 (the example is from Siegrist, n.d.).

• P(A) = 6/36 = 1/6, P(B) = 6/36 = 1/6, P(C) = 6/36 = 1/6 (the sum is 7 for (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)).
• A and B: the only outcome is (3,4), so P(A and B) = 1/36 = (1/6)(1/6). Independent.
• A and C: first die 3 and sum 7 forces the second die to be 4, so the only outcome is (3,4) and P(A and C) = 1/36 = (1/6)(1/6). Independent.
• B and C: second die 4 and sum 7 forces the first die to be 3, so P(B and C) = 1/36 = (1/6)(1/6). Independent.
• All three: the only outcome is (3,4), so P(A and B and C) = 1/36. The product is (1/6)(1/6)(1/6) = 1/216.

Since 1/36 ≠ 1/216, the triple condition fails. The events are pairwise independent but not mutually independent. The reason is visible in the sets: once A and B both occur, C is certain, because A and B together are contained in C.

Interpretation: what the result tells you

What you find	What follows
Every subset satisfies the product rule	The events are mutually independent.
Events are mutually independent	They are pairwise independent too, because every pair is a subset of the collection.
Every pair satisfies the product rule but the triple does not	Pairwise independent only; the events are not mutually independent (the dice example).
Any one subset fails the product rule	The events are not mutually independent.

Is mutual independence the same as mutually exclusive?

No. Independent and mutually exclusive do not mean the same thing (Illowsky and Dean, n.d.). Mutually exclusive events cannot occur together; independent events do not change each other’s probability. The two-event distinction is covered in independent vs mutually exclusive events, and non-overlap in disjoint events.

Assumptions and common mistakes

• Checking pairs only. Passing all pairwise tests does not establish mutual independence, as the dice example shows.
• Skipping subsets. With four events, the four triples and the single quadruple must be checked as well as the six pairs (11 conditions in all).
• Confusing independence with exclusivity. Two events with positive probability that cannot co-occur have an intersection of probability zero, while the product of their probabilities is positive, so they are not independent.
• Assuming it from context. Independence is a property of the probability model, so it is either built into the model’s definition or has to be checked against it.

Once events are mutually independent, joint probabilities are products; see the joint probability formula. Where events depend on each other, use conditional probability and Bayes’ theorem.

How to report it

State the assumption in the methods or model description and say which events it covers. Example wording:

We treated the three screening outcomes as mutually independent, so the probability of all three occurring was the product of their individual probabilities, P(A ∩ B ∩ C) = P(A)P(B)P(C).

If you tested it, report each probability you compared. Cite the textbook behind the definition; APA-style references are listed under Sources.

How to calculate it in Excel, R or Python

Each snippet enumerates the 36 dice outcomes and compares P(A and B and C) with P(A)P(B)P(C). The Python and R versions return 1/36 and 1/216.

Python

from itertools import product
from fractions import Fraction as F
o = list(product(range(1, 7), repeat=2))
A = {x for x in o if x[0] == 3}
B = {x for x in o if x[1] == 4}
C = {x for x in o if sum(x) == 7}
p = lambda s: F(len(s), 36)
print(p(A.intersection(B, C)), p(A) * p(B) * p(C))   # 1/36 1/216

R

d <- expand.grid(d1 = 1:6, d2 = 1:6)
A <- d$d1 == 3; B <- d$d2 == 4; C <- d$d1 + d$d2 == 7
mean(A * B * C)              # 1/36
mean(A) * mean(B) * mean(C)  # 1/216

Excel

List all 36 outcomes with the first die in A2:A37 and the second in B2:B37. Then:

P(A and B and C):   =SUMPRODUCT((A2:A37=3)*(B2:B37=4)*((A2:A37+B2:B37)=7))/36
P(A) x P(B) x P(C): =(COUNTIF(A2:A37,3)/36)*(COUNTIF(B2:B37,4)/36)*(SUMPRODUCT(--((A2:A37+B2:B37)=7))/36)

The first formula returns 0.02778 (1/36) and the second 0.00463 (1/216), so the triple condition fails.

FAQ

What does mutually independent mean?

The probability of every combination of the events occurring together equals the product of their individual probabilities, for every sub-collection (Siegrist, n.d.).

Does pairwise independence imply mutual independence?

No. In the two-dice example every pair is independent, yet P(A and B and C) = 1/36 while P(A)P(B)P(C) = 1/216. The reverse does hold: mutual independence implies pairwise independence.

How many conditions must be checked for n events?

2ⁿ − n − 1, the number of subsets with at least two events: 4 for three events and 11 for four (Siegrist, n.d.).

Is mutual independence the same as mutually exclusive?

No; they are different properties (Illowsky and Dean, n.d.).

Sources

• Illowsky, B., and Dean, S. (n.d.). Independent and mutually exclusive events. In Introductory statistics 2e (sec. 3.2). OpenStax. openstax.org, Independent and mutually exclusive events (date checked: 2026-10-29)
• Siegrist, K. (n.d.). Independence. In Probability, mathematical statistics, and stochastic processes (sec. 2.5). LibreTexts. stats.libretexts.org, Independence (date checked: 2026-10-29)