Independence and mutual exclusivity: Question bank

This is a question bank for the parent topic Independence and mutual exclusivity. Every item below hunts a specific misconception or boundary case. Read the prompt, think first, then reveal the answer. The answers are short but they always explain the why — no bare yes/no.

Before you start, keep three anchor facts in your head. These are the whole topic compressed:

Recall The three anchors everything rests on

Mutually exclusive means the events cannot happen together: $P (A \cap B) = 0$ .
Independent means one carries no information about the other: $P (A \cap B) = P (A) P (B)$ , equivalently $P (A ∣ B) = P (A)$ (valid only when $P (B) > 0$ ).
These are opposite ideas: mutual exclusivity is maximum dependence, independence is zero dependence.

The picture below is the whole topic in one glance — a Venn diagram. Each closed region is an event (a set of outcomes); the big box is the sample space $S$ (every possible outcome). Mutually exclusive events are two circles that do not touch; independent events are circles whose overlap area is exactly the product of their areas.

Figure — Independence and mutual exclusivity

A few symbols appear throughout — let's pin them down on that same picture before using them:

Two companion formulas complete the "one-glance" picture — the union of each kind of pair, and the fact that independence quietly spreads to complements:

Prerequisites if any line confuses you: 1.3.01-Basic-probability-concepts, 1.3.02-Conditional-probability. Payoff downstream: 3.1.01-Naive-Bayes-classifier and 5.2.03-Probabilistic-graphical-models.

The four logical combinations of "mutually exclusive?" × "independent?" fit into one 2×2 chart — keep it beside you as a sanity check:

True or false — justify

Two disjoint events with positive probability can also be independent.

False. Disjoint means

P (A \cap B) = 0

, independence needs

P (A \cap B) = P (A) P (B) > 0

— those clash, so non-trivial mutually exclusive events are always dependent.

A

and

B

are independent, then

P (A \cap B)

is always positive.

False. It is positive only when both

P (A) > 0

and

P (B) > 0

; if either event has probability

0

, the product is

0

and they are still (trivially) independent.

A

and

B

are independent, then

A

and

B^{c}

are independent too.

True.

P (A \cap B^{c}) = P (A) - P (A \cap B) = P (A) (1 - P (B)) = P (A) P (B^{c})

; independence passes to complements, and by the same algebra

A^{c} ⊥ B

and

A^{c} ⊥ B^{c}

hold as well.

An event

A

is always independent of the impossible event

\emptyset

True.

P (A \cap \emptyset) = 0 = P (A) \cdot 0 = P (A) P (\emptyset)

, so the independence equation holds trivially for the empty event.

An event

A

is always independent of the whole sample space

S

True.

P (A \cap S) = P (A) = P (A) \cdot 1 = P (A) P (S)

, so certain events give no information — they are independent of everything.

"Complementary events are independent" is a valid rule.

False.

A

and its complement

A^{c}

are mutually exclusive and exhaustive, so (unless one has probability

0

) they are maximally dependent, not independent.

P (A ∣ B) = P (A)

then also

P (B ∣ A) = P (B)

True (when

P (A) > 0

and

P (B) > 0

so both conditionals are defined). Independence is symmetric — both reduce to the single equation

P (A \cap B) = P (A) P (B)

, so if one direction holds the other does too.

Conditional independence given

C

implies plain (unconditional) independence.

False. They are logically separate;

A, B

can be dependent overall yet become independent once

C

is known (the flu-cough-fever example), and vice versa.

If two events are mutually exclusive, then

P (A ∣ B) = 0

True, provided $P (B) > 0$ (else

P (A ∣ B)

is undefined). Given

B

occurred,

A

is impossible, so

P (A ∣ B) = P (A \cap B) / P (B) = 0/ P (B) = 0

— knowing

B

rules out

A

Spot the error

"They're independent processes, so they can't both turn out heads."

Error: independence is informational separation, not physical separation. Both coins can land heads together; independence just means

P (H H) = P (H) P (H) = \frac{1}{4} > 0

A

and

B

have equal probability and are mutually exclusive, so they're independent."

Error: let

p

denote that shared value, i.e.

P (A) = P (B) = p

. Mutually exclusive gives

P (A \cap B) = 0

, but independence would need

P (A) P (B) = p^{2} > 0

(for

p > 0

) — a contradiction, not a match.

P (A ∣ B) \neq = P (A)

, therefore

A

and

B

are mutually exclusive."

Error: (reading

P (A ∣ B)

with its side-condition

P (B) > 0

) unequal conditional probability only means dependence, which is a whole spectrum. Mutual exclusivity (

P (A \cap B) = 0

) is one extreme; positive correlation (rain and clouds) is dependent but not exclusive.

P (A \cup B) = P (A) + P (B)

, so

A

and

B

must be independent."

Error: that additive formula is the mutually exclusive rule (overlap term

= 0

); the independent union is

P (A \cup B) = P (A) + P (B) - P (A) P (B)

, which keeps a subtraction — you've matched the wrong concept.

A

and

B

can't be independent because both happening is a rare coincidence."

Error: rarity is not the test. Even if

P (A \cap B)

is tiny, they are independent as long as that tiny value equals

P (A) P (B)

exactly — the equation, not the size, decides.

"For Naive Bayes we assume the features never occur together."

Error: Naive Bayes assumes features are conditionally independent given the class, not mutually exclusive. Features routinely co-occur; the assumption only says one feature adds no info about another once the class is fixed. See 3.1.01-Naive-Bayes-classifier.

"Since

A

and

B

are dependent, learning

B

must make

A

more likely."

Error: dependence can be either direction. Reading

P (A ∣ B)

under

P (B) > 0

, it could rise (positive) or fall (negative) relative to

P (A)

; dependence only says it changes, not that it increases.

Why questions

Why does the union formula simplify to

P (A) + P (B)

for mutually exclusive events?

Because the general rule

P (A \cup B) = P (A) + P (B) - P (A \cap B)

has an overlap term that vanishes when

P (A \cap B) = 0

; for independent events that overlap term is instead

P (A) P (B)

, giving

P (A) + P (B) - P (A) P (B)

Why do we say mutual exclusivity is "maximum dependence"?

Because knowing

A

occurred completely determines that

B

did not (with

P (A) > 0

P (B ∣ A) = 0

) — one event's outcome fixes the other's, the strongest possible information transfer.

Why is

P (A \cap B) = P (A) P (B)

the definition of independence rather than

P (A ∣ B) = P (A)

The product form works even when

P (B) = 0

(where the conditional is undefined), so it is the more general, always-valid statement; the conditional form is the same idea only when

P (B) > 0

. See 1.3.02-Conditional-probability.

Why does independence automatically carry over to the complements

B^{c}

and

A^{c}

Because

P (A \cap B^{c}) = P (A) - P (A \cap B) = P (A) - P (A) P (B) = P (A) P (B^{c})

— the algebra just factors out. If

B

carries no information about

A

, then "not

B

" carries none either, so all four pairs

A ⊥ B

A ⊥ B^{c}

A^{c} ⊥ B

A^{c} ⊥ B^{c}

stand or fall together.

Why can conditioning on a cause

C

make two effects independent?

C

(e.g. flu) is the common cause producing both symptoms, then once

C

is known the symptoms no longer signal each other — the shared explanation has been "used up," so

P (A \cap B ∣ C) = P (A ∣ C) P (B ∣ C)

. The labelled cause-and-effect graph below walks through this "common cause" wiring node by node.

Why does Naive Bayes multiply the per-feature likelihoods instead of estimating the full joint?

Conditional independence given the class lets

P (x_{1}, \dots, x_{n} ∣ y) = \prod_{i} P (x_{i} ∣ y)

, turning an exponentially large joint table into a handful of simple factors — that is what makes the classifier tractable.

Why isn't "unrelated in everyday language" a safe test for independence?

Intuition tracks causal relatedness, but independence is a precise numeric equality of probabilities; events can feel related yet be independent (die-1 and sum patterns) or feel unrelated yet be dependent, so only the equation decides.

The flu example only becomes concrete with numbers. The table below shows how the same two symptoms are dependent overall but independent once you fix the flu status.

Edge cases

Are a probability-zero event

A

and any event

B

independent?

Yes, trivially.

P (A \cap B) \leq P (A) = 0

P (A \cap B) = 0 = 0 \cdot P (B) = P (A) P (B)

— a null event carries and receives no information.

Can an event be independent of itself?

Only if $P (A) \in {0, 1}$ . Independence with itself needs

P (A) = P (A)^{2}

, whose only solutions are

0

and

1

— i.e. events that are certain or impossible.

Is every pair of mutually exclusive events also exhaustive (covering all outcomes)?

No. Rolling "a 2" and "a 5" are mutually exclusive but leave outcomes 1, 3, 4, 6 uncovered; exclusivity only forbids overlap, it does not require filling the sample space

S

If three events are pairwise independent, are they mutually (jointly) independent?

Not necessarily. Pairwise independence checks each pair, but joint independence also demands

P (A \cap B \cap C) = P (A) P (B) P (C)

; classic dice constructions satisfy every pair yet fail the triple, which matters in 5.2.03-Probabilistic-graphical-models.

What happens to the "opposites" claim when one event has probability zero?

It breaks gracefully. A zero-probability event is both mutually exclusive with and independent of the other — the two properties only conflict when both events have strictly positive probability.

Can two events be simultaneously mutually exclusive and independent?

Yes, but only in the degenerate case where at least one has probability

0

— then

P (A \cap B) = 0

equals

P (A) P (B) = 0

, satisfying both definitions at once. This is the "both" corner of the 2×2 chart above.