Unit 8 · Statistics & ProbabilityS4 · Probability · Telemetry

Probability & Venn Diagrams

Count outcomes, complements, unions, intersections, and independent events.

28 min
Unlocks: Risk Assessment

Ready when you are

7 steps · about 28 minutes

Start with the warm-up.

Start

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Warm-up

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Key terms

complement: All outcomes outside an event.
intersection: Outcomes in both events.
union: Outcomes in at least one event.

Start by naming the quantity each problem asks for. That habit blocks the most common ACT trap.

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Try it first

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Take a calm first shot. Label what you know, write one relationship, and make the best choice you can.

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The big idea

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Build the picture before the arithmetic

A Venn overlap belongs to both circles but counts once in the union. Independent events multiply. Mutually exclusive events have no overlap; they are usually not independent.

Key idea

The decision that matters

Probability is favorable outcomes over equally likely total outcomes. For “at least one,” the complement is often fastest.

Addition rule

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

Write this relationship before inserting numbers. It keeps labels and units attached to the work.

A compact example

If $P(A)=0.4$ , $P(B)=0.5$ , and $P(A\cap B)=0.2$ , then $P(A\cup B)=0.4+0.5-0.2=0.7$ .

Cue	Best move
At least one	try the complement
Overlapping groups	subtract the intersection once

Trap

Counting the overlap twice

This error produces a believable answer. Pause before computing and say what each number means.

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Worked examples

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Skip check · 2 questions

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Worked example

Example 1: Use inclusion-exclusion

Problem

Of 60 students, 28 study Spanish, 25 study French, and 9 study both. How many study at least one of the two languages?

Adding 28 and 25 counts the 9 students in both groups twice.

Subtract one copy of the overlap: $28+25-9=44$ .

How should the overlap be handled when counting the union?

So 44 students study at least one language.

Why is the overlap subtracted once?

Worked example

Example 2: Use a complement

Problem

A fair die is rolled twice. What is the probability of getting at least one 6?

The complement is no 6 on either roll.

$P(\text{no 6})=(5/6)^2=25/36$ .

Which setup finds the probability of at least one 6?

Therefore $P(\text{at least one 6})=1-25/36=11/36$ .

Why does the complement method work here?

Faded example · you finish it

Example 3: You finish it

Problem

P(A)=0.55

P(B)=0.40

, and

P(A\cap B)=0.15

, find

P(A\cup B)

Use

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

Which completion is correct?

Which reason validates the result?

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Practice

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Spot the mistake

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Spot the mistake

Where did it go wrong?

One line has an error. Click the line where it first goes wrong.

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Exit ticket

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Optional · 4 questions · Hardest levelChallenge set: the kind of question that decides a 36

These are last-ten-questions hard. Take your time, use Desmos when it helps, and expect to miss some: that's where the learning is.