Math // Probability

Probabilities Sum to One

Understand that when all possible outcomes of a trial are listed, their probabilities must add up to 1; use this to find the probability of an event NOT happening: P(not A) = 1 − P(A); apply this shortcut to avoid counting all unfavourable outcomes directly

Ages
10 to 11
Type
Concept
Builds on
2 topics
Leads to
1 topic

Quick check

If the probability of it raining tomorrow is 0.3, can the learner work out the probability of it NOT raining — and explain why all probabilities in a situation must add up to 1?

Ask this to see whether the topic is secure.

Topic // Mastery

What mastery looks like

What a learner can do once this topic is secure.
  • List all outcomes of spinning a 4-colour spinner and verify their probabilities add up to 1
  • Calculate P(not rolling a 3) as 1 − 1/6 = 5/6 using the complement rule
  • Spot an error in a probability table where the values don't sum to 1 and explain what's wrong

Topic // Map

Where this fits

What to learn first and what this topic opens up, with the reason for each link. Required links come first.
  • RequiredMust come first
  • HelpfulHelps, but not essential

Before this

2 topics
  • The 0-to-1 Probability ScaleRequired

    Probability // Ages 10 to 11

    The complement rule P(not A) = 1 − P(A) requires understanding probability as a number that lies between 0 and 1

  • Calculating Simple ProbabilityHelpful

    Probability // Ages 10 to 11

    Using the complement rule is easier once students can calculate basic probabilities and see that favourable + unfavourable outcomes cover all possibilities

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Probabilities Sum to One

Ages 10 to 11 // Concept

Leads to

1 topic
  • Complementary eventsRequired

    Probability // Ages 11 to 13

    The formal mutually-exclusive outcomes rule extends the simpler complement rule introduced at age 10-11

Topic // Research-informed

Teaching notes

From the Marble Skill Taxonomy. These notes have not been validated in classrooms, so treat them as a starting point for your own judgement.

The approach

Teach that when you list ALL the possible outcomes of a trial — with no overlaps and nothing left out — their probabilities must total exactly 1. This lets a learner find a missing probability, or the probability of an event NOT happening, by subtracting from 1 (P(not A) = 1 − P(A)) instead of counting every unfavourable outcome. The reasoning only holds when the outcomes are mutually exclusive (no overlap) and exhaustive (cover everything).

Common misconceptions

  • Ignoring the sum-to-1 constraint and assigning a missing outcome an arbitrary or 'equal' value by analogy to another
  • Treating 'not A' as a single other outcome rather than everything except A, or thinking P(not A) equals P(A)
  • Adding probabilities of overlapping or non-exhaustive events and still expecting them to total 1
  • Adding fractions without a common denominator when totalling probabilities

Useful representations

  • Unit bar (length 1) partitioned into outcome segments
  • Complete outcome table with probabilities totalling 1
  • P(not A) shown as the remaining part of the whole
  • Spinner whose sectors fill the entire disc

Teaching practices

  • Model the whole of possibility as a bar of length 1 partitioned among the outcomes, so 'the rest of the bar' IS the complement
  • Establish that the list must be exhaustive (covers everything) and mutually exclusive (no overlaps) before the sum-to-1 argument applies
  • Teach P(not A) = 1 − P(A) as a shortcut, contrasting it with the slower count-all-the-others method on the same problem
  • Use spot-the-error tables whose probabilities don't total 1, and ask what must be wrong
  • Link back to calculating simple probabilities so children see favourable + unfavourable together exhaust the outcomes

Activities to try

  1. 1

    Fill the unit bar: the leftover IS the missing probability

    Why it works: The whole of possibility has length 1; once some outcomes are placed, the leftover segment IS the missing probability, and 'not red' is the whole rest of the bar — everything except red — not a single other colour.

    1. Take a bar exactly one unit long that stands for everything that can happen when you draw a counter from a bag of red, blue and green counters.
    2. Set red to one-half of the bar and blue to one-sixth, marking those segments.
    3. Before being told, work out what must fill the remaining strip and state it: green = 1 − (1/2 + 1/6) = 1 − 2/3 = 1/3.
    4. Mark 'P(not red)' by highlighting the entire rest of the bar (blue and green together) and give its value as 1 − P(red) = 1/2.
    5. State in words why green could NOT be a free choice and why 'green must be 1/6 like blue' is wrong.
  2. 2

    Spinner that fills the whole disc

    Why it works: A spinner's sectors are another model of the whole-equals-1 idea: because the sectors must fill the disc, the last sector's size (and probability) is forced by the others.

    1. Look at (or draw) a spinner whose coloured sectors must fill the entire disc with no gap and no overlap.
    2. Given red = one-half and blue = one-third, predict the remaining sector's probability (yellow) before it is revealed: 1 − (1/2 + 1/3) = 1/6.
    3. Check the revealed sector really is a sixth of the disc.
    4. Tap or mark each sector in turn and track the running total as it reaches exactly 1.
  3. 3

    Complement shortcut versus counting the others

    Why it works: P(not A) = 1 − P(A) is a shortcut for 'everything except A', and it equals the sum of all the other outcomes because together they fill the whole; the same subtraction works when the known part is a compound event like 'red or blue'.

    1. Calculate P(not rolling a 3) on a fair six-sided die two ways: first by subtracting, 1 − 1/6 = 5/6; then by counting all the other outcomes (1,2,4,5,6) and adding their probabilities (five sixths).
    2. Compare the two methods, note they give the same answer, and that subtracting from 1 is faster when there are many unfavourable outcomes.
    3. For the counter bag, find P(not red) as 1 − P(red) = 1/2 and confirm it matches blue + green (1/6 + 1/3 = 1/2).
    4. Extend to a compound complement: a different bag of red, blue and green counters has P(red or blue) = 3/4; find P(green) = 1 − 3/4 = 1/4, since each counter is exactly one of the three colours and the three colours exhaust the bag.

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