Probability · KS3

Mutually Exclusive Outcomes

When you spin a spinner, exactly one thing happens: it lands on one colour, never two at once. That simple fact gives one of the most useful rules in probability: if you list every possible outcome, and no two of them can happen together, their probabilities add up to exactly 1. On this page you'll learn the words for this, see why it works, and use it to find probabilities nobody has told you.

National curriculum KS3: apply the property that the probabilities of an exhaustive set of outcomes sum to one; apply the property that the probabilities of an exhaustive set of mutually exclusive events sum to one

1 · Can both happen at the same time?

Some things simply can't happen together. A traffic light can't be on red and on green at the same moment. You can't be in London and in Paris at the same time. When you flip a coin once it lands on heads or tails, never both.

First, two words. An outcome is one possible result, like "the dice shows 4". An event is an outcome or a group of outcomes you are interested in, like "the dice shows an even number" (that's the outcomes 2, 4 and 6). If you need a reminder of how to work out a probability, such as P(even) = 3 ÷ 6, see the probability scale page.

Two events are mutually exclusive if they cannot happen at the same time. "Mutually" means "each other", and "exclude" means "keep out": each event keeps the other one out.

The test: try to find one single outcome that is in both events. If you can find even one, the events are not mutually exclusive. If there is no such outcome, they are.

Have a go. Decide for each pair, then the diagram shows any overlap, outlined in orange. Outcomes in neither event fade out.

2 · Exhaustive: nothing left out

A list of events is exhaustive if it covers every possible outcome, so that one of them is certain to happen. ("Exhaustive" is like "exhausting" all the possibilities: you've used them all up.)

On one dice the outcomes are 1, 2, 3, 4, 5, 6. The events "even" and "odd" are exhaustive: every number is one or the other. But "less than 3" and "more than 3" are not exhaustive, because 3 has been left out.

The most useful lists are both: every outcome is in exactly one event. Think of sorting socks into drawers: every sock goes in a drawer (exhaustive) and no sock is in two drawers at once (mutually exclusive).

Pick events below to cover the dice. Faces covered once turn teal, faces covered twice or more turn orange (overlap), and faces you have missed stay dashed grey. Can you find three different sets that are both exhaustive and mutually exclusive?

3 · Sharing out the 1

A probability of 1 means certain. If your list of outcomes is exhaustive, one of them is certain to happen, so together they have probability 1. If the outcomes are also mutually exclusive, none of them overlap, so that 1 is shared out between them with nothing counted twice, like slices of one cake that together make the whole cake.

Concrete example first: a fair dice has six outcomes, each with probability 1/6. Add them: 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1.

Now drag the white handles on the spinner (or on the bar underneath) to change the sizes of the sectors. However you share it out, the four probabilities always add to exactly 1, because the bar is always exactly 1 long and is always completely filled.

P(A) + P(B) + P(C) + … = 1

…as long as the outcomes A, B, C, … are mutually exclusive and exhaustive. The same is true of results from experiments: the relative frequencies of all the outcomes add up to 1 too (see theoretical and experimental probability).

4 · The chance it doesn't happen: P(not A)

A weather app says there is a 30% chance of rain. What is the chance of no rain? You probably already know: 70%. As probabilities that is 0.3 and 0.7, and notice that 0.3 + 0.7 = 1.

For any event A, the event "not A" means "A does not happen". A and not A can't both happen (mutually exclusive) and one of them must happen (exhaustive). So they share out the 1:

P(A) + P(not A) = 1, so P(not A) = 1 − P(A)

The grid has 100 squares and stands for the whole probability, 1. Each square is 0.01. Slide to shade event A in teal; whatever is left over, in grey, is "not A".

Try it

You can type decimals like 0.65 or fractions like 5/8.

5 · Finding a missing probability in a table

Exam questions often give a table of probabilities for every outcome with one missing. Because the outcomes are mutually exclusive and exhaustive, the full list must add up to 1. So:

  1. Add up the probabilities you know.
  2. Take that total away from 1. What's left is the missing probability.
  3. Check: all of them together should now make exactly 1.

Example: P(red) = 0.2, P(blue) = 0.5, P(green) = ?. Known total 0.2 + 0.5 = 0.7, so P(green) = 1 − 0.7 = 0.3. Check: 0.2 + 0.5 + 0.3 = 1 ✓

With an unknown x. Sometimes the missing probabilities are written with a letter, such as 0.2, 0.35, 2x and x. Write "everything adds to 1" as an equation and solve it: 0.2 + 0.35 + 2x + x = 1, so 0.55 + 3x = 1, so 3x = 0.45, so x = 0.15. Notice 2x + x is 3x, so you divide by 3 at the end.

Practise below. The bar shows the known probabilities in teal and the unknown part in purple. If you go wrong, the feedback tries to spot which mistake you made.

6 · The OR rule: P(A or B) = P(A) + P(B)

A bag has 10 counters: 3 red, 2 blue and 5 green. What is the probability of taking a counter that is red or blue? Count them: 3 + 2 = 5 of the 10 counters, so 5/10. And look: P(red) + P(blue) = 3/10 + 2/10 = 5/10. Adding works, because no counter is both red and blue.

If A and B are mutually exclusive:
P(A or B) = P(A) + P(B)

But if the events overlap, adding counts the overlapping outcomes twice and gives an answer that is too big. Choose two events on the spinner below and compare "add the probabilities" with "count the sectors".

Only add when the events can't happen together. When they overlap, list or count the outcomes instead. Diagrams that show overlapping events clearly, called Venn diagrams, are on the sample spaces and Venn diagrams page.

7 · Common mistakes and how to spot them

Spot the mistake in each student's working. The bar shows their numbers on a scale from 0 to 1; anything past 1 shows in orange.

Worked examples

A spinner can land on red, blue or green. P(red) = 0.3 and P(blue) = 0.45. Find P(green).
Can I use "add to 1"? The spinner lands on exactly one colour (mutually exclusive) and these are the only colours (exhaustive). Yes.
Add the ones I know: 0.3 + 0.45 = 0.75.
Take away from 1: P(green) = 1 − 0.75 = 0.25.
Check: 0.3 + 0.45 + 0.25 = 1 ✓ (0.25 is the same as 1/4.)
The probability a train is late is 0.08. Find the probability it is not late. Then: P(Sam wins) = 3/7; find P(Sam does not win).
Late and not late are mutually exclusive and exhaustive, so P(not late) = 1 − P(late).
P(not late) = 1 − 0.08 = 0.92. Check: 0.08 + 0.92 = 1 ✓
With fractions, write 1 as a fraction with the same denominator: 1 = 7/7.
P(Sam does not win) = 7/7 − 3/7 = 4/7. Check: 3/7 + 4/7 = 7/7 = 1 ✓
The table shows probabilities for a bag of counters: red 0.2, blue 0.35, green 2x, yellow x. Find x and P(green).
Everything adds to 1: 0.2 + 0.35 + 2x + x = 1.
Collect like terms: 0.2 + 0.35 = 0.55 and 2x + x = 3x, so 0.55 + 3x = 1.
Subtract 0.55 from both sides: 3x = 1 − 0.55 = 0.45.
Divide both sides by 3: x = 0.45 ÷ 3 = 0.15.
P(green) = 2x = 2 × 0.15 = 0.3.
Check: 0.2 + 0.35 + 0.3 + 0.15 = 1 ✓
Outcomes A, B and C are mutually exclusive and exhaustive. P(A) = 1/4 and P(B) = 1/3. Find P(C).
Common denominator: 12 works for both 4 and 3. 1/4 = 3/12 and 1/3 = 4/12.
Add: 3/12 + 4/12 = 7/12.
Take away from 1 = 12/12: P(C) = 12/12 − 7/12 = 5/12.
Check: 3/12 + 4/12 + 5/12 = 12/12 = 1 ✓
A counter is taken from a bag. P(red) = 0.15 and P(blue) = 0.4. Find P(red or blue) and P(neither red nor blue).
Mutually exclusive? One counter can't be red and blue, so yes: use the OR rule.
P(red or blue) = 0.15 + 0.4 = 0.55.
"Neither" is "not (red or blue)": P = 1 − 0.55 = 0.45.
A fair dice is rolled. Find P(even or more than 4).
List each event: even = 2, 4, 6 (3/6); more than 4 = 5, 6 (2/6).
Overlap? 6 is in both, so they are not mutually exclusive. Adding 3/6 + 2/6 = 5/6 would count the 6 twice: wrong.
Count instead: the outcomes that are even or more than 4 are 2, 4, 5, 6: four outcomes.
P(even or more than 4) = 4/6 = 2/3.

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