GCSE · Maths · AQA · Spec 8300

Exhaustive sets and mutually exclusive events sum to 1

Add up the probabilities of a list of events and you'd expect 1. Sometimes you get 5/6, sometimes 1.1, and sometimes that's fine. Which lists HAVE to make 1?

Probability · one roll of a dice

Do these two events cover the dice exactly once?

Roll a fair dice once. Event A is 'an even number'. Event B is 'a number less than 3'. Place each score: it can go in A, in B, in both, or in neither.

A Even
B Less than 3

What do you think?

Is 1.1 a mistake?

A student is picked at random from a class. The probability that they have a sister is 0.6. The probability that they have a brother is 0.5. Add those together and you get 1.1.

Which of these is closest to what you think?

Something's wrong. Probabilities have to add up to 1, so one of those numbers must be a mistake.They should add to 1, so the probability of having a brother ought to be 1 − 0.6 = 0.4.Nothing's wrong. A student could have a sister and a brother, or neither, so these two don't have to add to 1.A total over 1 is impossible, so 1.1 must be wrong. A total like 0.9 would have been fine.

How sure are you?

I'm sureI think soI'm guessing

Something's wrong. Probabilities have to add up to 1, so one of those numbers must be a mistake.

This is a common misconception

The add-to-1 rule isn't true for any old list. It only works when the events cover every possibility and nothing gets counted twice, and 'has a sister' and 'has a brother' fail both checks. A student with a sister AND a brother is counted in the 0.6 and again in the 0.5. A student with neither isn't counted at all. So nothing forces these two to make 1, and 1.1 can be exactly right.

They should add to 1, so the probability of having a brother ought to be 1 − 0.6 = 0.4.

This is a common misconception

1 − 0.6 = 0.4 is a real probability, but it's the chance of NOT having a sister. 'Has a sister' and 'doesn't have a sister' cover every student, and nobody can be both, so that pair does add to 1. 'Has a brother' is a different event altogether, and the subtraction tells you nothing about it.

Nothing's wrong. A student could have a sister and a brother, or neither, so these two don't have to add to 1.

This matches the evidence

Spot on. The rule needs a list that's exhaustive (it covers every possibility) and mutually exclusive (nothing overlaps). A student with both is in the overlap, and a student with neither is left outside. Either one is enough to stop the total being forced to 1.

A total over 1 is impossible, so 1.1 must be wrong. A total like 0.9 would have been fine.

This is a common misconception

One probability can't be bigger than 1, but a total of several can be, when the events overlap. Every student with a sister and a brother is counted in both numbers, and that double counting is what pushes 0.6 + 0.5 past 1. And a total like 0.9 wouldn't prove the list was fine: a total under 1 can still hide an overlap.

The rule belongs to the list, not to the numbers. Only when a list covers everything, with nothing counted twice, do its probabilities have to add to 1.

Run the check

Which lists have to add to 1?

Put each list of events where it belongs.

Still to sort

Must add to 1 (0)

Every outcome is covered, and none is in two events.

Leaves an outcome out (0)

No overlaps, but something that could happen isn't in any event.

Where the line is: A list with no overlaps can still fall short of 1. Check for gaps as well as overlaps.

Counts an outcome twice (0)

Everything is covered, but some outcome is in two events.

Where the line is: Covering every outcome isn't enough on its own. An overlap pushes the total above 1.

Both problems (0)

Something is left out AND something is counted twice.

7 of 7 still to sort.

Each list is the events for ONE trial. Think about every outcome that could happen, then decide what's wrong with the list, if anything.

Your turn

Find the missing probability

A biased spinner lands on red, blue, green or yellow. P(red) = 0.2 and P(blue) = 0.35. Yellow is twice as likely as green: if P(green) = x, then P(yellow) = 2x. Find P(green) and P(yellow).

Each spin lands on exactly one colour, and there are no other colours. So the four colours are exhaustive and mutually exclusive, and their probabilities add to 1.Check the list first. This one passes, so the rule is safe to use.
missing step

Which line is step 2?

0.2 + 0.35 + x + 2x = 1000.2 + 0.35 + x + 2x = 10.2 + 0.35 = x + 2x

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

Nothing left out, nothing counted twice: then the probabilities make exactly 1.

What you need to know

An outcome is one possible result of a trial, like scoring a 4 on one roll of a dice. An event is a group of outcomes, like 'an even number'.
A set of outcomes or events is exhaustive when together they cover everything that can happen. Nothing is left outside.
Events are mutually exclusive when no two of them can happen at the same time, so no outcome belongs to two of them. The different outcomes of a single trial are always mutually exclusive: one roll can't land on a 2 and a 5 at once.
The probabilities of an exhaustive set of outcomes add up to 1. The same goes for an exhaustive set of mutually exclusive events. So a missing probability = 1 − (the total of the others).
The simplest case is an event and its opposite. Every outcome is either A or not A, never both, so P(not A) = 1 − P(A).

The big picture

The probabilities of a set of events add up to exactly 1 when two things are true: together they cover every possible outcome (exhaustive), and no outcome belongs to more than one of them (mutually exclusive). When both hold, a missing probability is 1 minus the total of the ones you know.

Key points

1Guaranteed to add to 1 when nothing is left out (exhaustive) AND nothing is counted twice (mutually exclusive).

2An outcome left out pulls the total down; an outcome in two events pushes it up.

3Missing probability = 1 − (total of the known probabilities). With letters, write 'total = 1' and solve.

4P(not A) = 1 − P(A).

5Probabilities written as decimals or fractions add up to 1, not 100.

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Worked example

Problem

A jar holds only strawberry, lemon and lime sweets. One sweet is taken at random. P(strawberry) = 1/4 and P(lemon) = 5/12. (a) Find P(lime). (b) Find P(not lemon).

⚠ Watch out

Treating 'less than 3' and 'more than 3' as opposites. On a dice they are 1, 2 and 4, 5, 6, so the 3 belongs to neither, and their probabilities add to 5/6, not 1. When a list splits at a number, check where that number goes.

🧠

Memory hook

Nothing outside, nothing twice: then it's 1. Picture the Venn diagram. An empty outside and an empty overlap are your green light to use 'they add to 1'.

✓

Check yourself

A letter is picked at random from the word MATHS. Must P(a vowel) + P(a consonant) equal 1? What about P(a vowel) + P(a letter in the word HAT)? Give a reason each time.

Flashcards

(7)

What does 'exhaustive' mean for a set of outcomes or events?

Together they cover every possible outcome. Nothing that could happen is left out.

What does 'mutually exclusive' mean?

No two of the events can happen at the same time, so no outcome belongs to more than one of them.

When must the probabilities of a set of events add up to exactly 1?

When the events are exhaustive and mutually exclusive: every outcome is covered, and none is counted twice.

Why are the outcomes of a single trial always mutually exclusive?

One trial gives exactly one outcome. A single roll of a dice can't land on a 2 and a 5 at once.

How do you get P(not A) from P(A)?

P(not A) = 1 − P(A), because every outcome is either A or not A, and never both.

The probabilities of a list of events add up to more than 1. What does that tell you?

At least two of the events overlap: some outcome is being counted more than once.

Some mutually exclusive events have probabilities that add up to less than 1. What does that tell you?

They aren't exhaustive. Some outcome is left out, and the gap up to 1 is the probability of everything they miss.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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Do these two events cover the dice exactly once?

Is 1.1 a mistake?

Which lists have to add to 1?

Find the missing probability

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