M1–M2StatisticsCore

Independence and Tree Diagrams

Q: Can two events be both independent and mutually exclusive?

Only if one has probability zero. Exclusive events satisfy P(A B)=0, but independence requires P(A)P(B), which is non-zero when both are possible.

Two events are $\textbf{independent}$ when one happening tells you nothing about the other. That is a precise, testable condition — not something to assume because the events feel unrelated — and tree diagrams are how multi-stage problems get organised once you know whether it holds.

35 min Probability

The big picture

Independence is the assumption that makes the rest of statistics tractable. The binomial distribution requires independent trials; hypothesis tests assume independent observations; the multiplication rule $P(A)P(B)$ only works under it. So the habit worth building is checking rather than assuming — and recognising the situations where it plainly fails, such as sampling without replacement, where each draw changes what is left. Getting this right here is what stops the distributions chapter being applied to data it does not fit.

What you'll be able to do

State the condition for independence and test whether two events satisfy it
Distinguish independent from mutually exclusive events
Construct tree diagrams for with- and without-replacement problems
Use tree diagrams to answer "at least one" and conditional questions

The condition for independence

Events $A$ and $B$ are independent when knowing $B$ has happened leaves the probability of $A$ unchanged: $P(A \mid B) = P(A)$ . Substituting into the conditional formula gives the version usually used for testing.

The test is a calculation, not a judgement. Compute $P(A) \times P(B)$ and compare it with $P(A \cap B)$ ; if they are equal the events are independent, and if not they are not.

All three forms are equivalent, so any one of them can be quoted — but the multiplication form is easiest to check because it needs no division.

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

The test for independence.

P(A \mid B) = P(A)

Equivalent: conditioning changes nothing.

P(B \mid A) = P(B)

Independence is symmetric.

Example 1Given

P(A) = 0.4

P(B) = 0.5

and

P(A \cap B) = 0.2

, determine whether

A

and

B

are independent.

1Compute

P(A)\times P(B) = 0.4 \times 0.5 = 0.2

2Compare with

P(A\cap B) = 0.2

3They match.

AnswerIndependent, since

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

Example 2Now suppose

P(A\cap B) = 0.3

instead. Are they independent?

P(A)P(B) = 0.2

still.

0.3 \neq 0.2

AnswerNot independent — the events occur together more often than independence would predict.

Tip — Show the comparison explicitly. " $P(A)P(B) = 0.2 = P(A\cap B)$ , so independent" earns the mark; stating the conclusion alone does not.

Independent is not mutually exclusive

These two terms are routinely confused, and they are close to opposites. $\textbf{Mutually exclusive}$ means the events cannot both occur, so $P(A \cap B) = 0$ . $\textbf{Independent}$ means one occurring does not change the probability of the other.

If $A$ and $B$ are mutually exclusive and both have non-zero probability, then knowing $A$ occurred tells you $B$ definitely did $\textit{not}$ — which is a very large change in probability. So they cannot be independent.

Formally: mutually exclusive gives $P(A\cap B) = 0$ , while independence requires $P(A)P(B)$ , which is non-zero whenever both events are possible. The two conditions are incompatible unless one event has probability zero.

Why this works

A useful mental check: mutually exclusive events are maximally $\textit{dependent}$ . Learning that one happened resolves the other completely, which is the opposite of telling you nothing.

Tree diagrams

A tree diagram lays out a sequence of stages. Each branch carries a probability, the branches from any single point sum to 1, and the probability of a complete path is the product of the probabilities along it.

To find the probability of an event described in words, identify every path that satisfies it and add those path probabilities. Multiply along, add across — that is the whole method.

With $\textbf{replacement}$ , the second-stage probabilities match the first: the situation is restored between draws, so the stages are independent. $\textbf{Without replacement}$ , both the favourable count and the total fall, so the second-stage branches differ depending on what happened first.

P(\text{path}) = \prod \text{branch probabilities}

Multiply along a path.

P(\text{event}) = \sum \text{matching paths}

Add across paths.

Example 1A bag holds 4 red and 6 blue counters. Two are drawn without replacement. Find the probability that exactly one is red.

1Path red-then-blue:

\tfrac{4}{10} \times \tfrac{6}{9} = \tfrac{24}{90}

2Path blue-then-red:

\tfrac{6}{10} \times \tfrac{4}{9} = \tfrac{24}{90}

3Exactly one red means either path, so add them:

\tfrac{48}{90}

Answer

\tfrac{8}{15}

Tip — Both orders must be counted. "Exactly one" almost always means two paths, and omitting the second halves the answer — one of the most common errors in the topic.

"At least one", and reading conditionals off a tree

For "at least one", the complement is almost always faster. The opposite of at least one success is $\textbf{no}$ successes — a single path — so $P(\text{at least one}) = 1 - P(\text{none})$ .

Tree diagrams also answer conditional questions. To find $P(A \mid B)$ , add the path probabilities for paths satisfying both, then divide by the total probability of all paths satisfying $B$ . The denominator is the restricted sample space, exactly as in the previous lesson.

Example 1A component fails with probability

0.1

, independently each day. Find the probability it fails at least once in three days.

1Not failing on one day:

0.9

2Not failing on all three (independent, so multiply):

0.9^{3} = 0.729

3At least one failure:

1 - 0.729

Answer

0.271

Example 2Using the 4 red and 6 blue bag without replacement, find the probability the first was red given that exactly one red was drawn.

1Paths with exactly one red: red-blue

\tfrac{24}{90}

and blue-red

\tfrac{24}{90}

, totalling

\tfrac{48}{90}

2Of these, "first was red" is the red-blue path:

\tfrac{24}{90}

3Divide:

\tfrac{24/90}{48/90}

Answer

\tfrac{1}{2}

Why this works

Enumerating "at least one" directly for three days would need seven paths; the complement needs one. The saving grows with the number of stages, which is why this shortcut is worth reaching for automatically.

Think like an examiner

For independence, show $P(A)P(B)$ and $P(A\cap B)$ side by side and state whether they are equal.
Label every branch of a tree with its probability and check each set of branches sums to 1.
For without-replacement problems, show the reduced denominators at the second stage.
Use the complement for "at least one" and say that is what you are doing.

Common misconceptions

Treating mutually exclusive events as independent.

They are incompatible. Exclusive means

P(A\cap B)=0

; independence needs

P(A)P(B) \neq 0

Counting only one order for "exactly one".

Both orders are valid paths and must be added.

Reusing first-stage probabilities after drawing without replacement.

Both the favourable count and the total drop by one.

Assuming events are independent because they seem unrelated.

Independence is a numerical condition. Test it with

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

Independence and trees

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

Independence test.

P(A\mid B) = P(A)

Equivalent form.

P(\text{at least one}) = 1 - P(\text{none})

The complement shortcut.

P(\text{path}) = \prod \text{branches}

Multiply along, add across.

Stretch yourself

A box contains 3 faulty and 7 working bulbs. Two are tested without replacement. Find the probability that at least one is faulty, and then the probability that both are faulty given that at least one is.

Hint — Use the complement for the first part. For the second, the conditioning event is "at least one faulty".

Reveal solution

SolutionFor at least one faulty, use the complement.

P(\text{neither faulty}) = \tfrac{7}{10}\times\tfrac{6}{9} = \tfrac{42}{90} = \tfrac{7}{15}

, so

P(\text{at least one}) = 1 - \tfrac{7}{15} = \tfrac{8}{15}

. For the conditional part,

P(\text{both faulty}) = \tfrac{3}{10}\times\tfrac{2}{9} = \tfrac{6}{90} = \tfrac{1}{15}

. Conditioning on "at least one faulty" restricts the sample space to probability

\tfrac{8}{15}

, so

P(\text{both} \mid \text{at least one}) = \dfrac{1/15}{8/15} = \tfrac{1}{8}

. Note this is much larger than the unconditional

\tfrac{1}{15}

— learning that at least one is faulty rules out the many outcomes with none, which raises the relative weight of the both-faulty case.

\blacksquare

Questions students ask

Can two events be both independent and mutually exclusive?

Only if one has probability zero. Exclusive events satisfy

P(A\cap B)=0

, but independence requires

P(A)P(B)

, which is non-zero when both are possible.

Is sampling without replacement ever independent?

No. Removing an item changes the composition for the next draw, so each stage depends on what came before. Sampling with replacement is independent.

How do I know which paths to add?

Translate the description into outcomes first. "Exactly one" means one success and one failure in either order; "at least one" means every path except all-failures.

Do branch probabilities always sum to 1?

Yes, at every branching point, because the branches cover all possibilities from that point. It is a quick and reliable check on a completed tree.

Key takeaways

Independence means $P(A\cap B) = P(A)P(B)$ — a test, not an assumption.
Mutually exclusive and independent are incompatible for events of non-zero probability.
On a tree, multiply along a path and add across paths.
Without replacement changes both the numerator and denominator at the next stage.
"At least one" is fastest via the complement: $1 - P(\text{none})$ .