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Year 10 · Probability · 18 min

Conditional language and restricted sample spaces

Restrict the sample space to the given information before finding a conditional probability.

01 / LEARN

The idea, step by step

The condition changes the population

P(A|B) means the probability of A given that B has occurred. Keep only outcomes in B, then count those also in A. The denominator is the size or probability of B, not the whole original sample space.

For a fair die known to be greater than 3, only 4,5,6 remain. Two of these are even, so the conditional probability is 2/3 rather than the original 1/2.

P⁢(A∣B)=P⁢(A∩B)P⁢(B),P⁢(B)>0

Read the order of the events

P(girl|walk) and P(walk|girl) ask different questions. In the first, look only at walkers. In the second, look only at girls. The overlap count can be the same while the denominators differ.

Check that the condition is possible. A conditional probability given an event of probability zero is not found by dividing by zero. In finite counting questions, the restricted group must contain at least one outcome.

Invented survey of 60 students; every student is in exactly one travel category.
TravelGirlBoyTotal
Cycle12820
Walk182240
Total303060

A common trap

Dividing the overlap count by the whole total after a condition is given.

Instead: Divide by the count in the given group.

02 / WORKED EXAMPLES

See the thinking

Example 1

Restrict a die result

A fair die result is known to be odd. Find the probability it is prime.

  1. Keep {1,3,5}.
  2. The prime outcomes are 3 and 5.
2/3.
Example 2

A survey condition

Using the supplied table, find P(walk|girl).

Invented survey of 60 students; every student is in exactly one travel category.
TravelGirlBoyTotal
Cycle12820
Walk182240
Total303060
  1. There are 30 girls.
  2. 18 of those girls walk.
  3. 18/30=3/5.
3/5.

Check your understanding

Using the table, find P(girl|walk).

Show answer

9/20.

18 of the 40 walkers are girls, so 18/40=9/20.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Read the questions online
  1. A fair six-sided die result is known to exceed 3. Find P(even).
    2 marks
  2. A fair six-sided die result is known to be even. Find P(greater than 3).
    2 marks
  3. Find P(cycle|girl) from the table.
    2 marks
    Invented survey of 60 students; every student is in exactly one travel category.
    TravelGirlBoyTotal
    Cycle12820
    Walk182240
    Total303060
  4. Find P(girl|cycle) from the table.
    2 marks
    Invented survey of 60 students; every student is in exactly one travel category.
    TravelGirlBoyTotal
    Cycle12820
    Walk182240
    Total303060
  5. Using the travel table, explain why P(cycle|girl) and P(girl|cycle) have different denominators.
    3 marks
    Invented survey of 60 students; every student is in exactly one travel category.
    TravelGirlBoyTotal
    Cycle12820
    Walk182240
    Total303060
  6. P(A and B)=0.12 and P(B)=0.3. Find P(A|B).
    2 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. A fair six-sided die result is known to exceed 3. Find P(even).
    2 marks

    Restrict to 4,5,6;4 and 6 are even.

    2/3.
  2. A fair six-sided die result is known to be even. Find P(greater than 3).
    2 marks

    Restrict to 2,4,6;4 and 6 exceed 3.

    2/3.
  3. Find P(cycle|girl) from the table.
    2 marks
    Invented survey of 60 students; every student is in exactly one travel category.
    TravelGirlBoyTotal
    Cycle12820
    Walk182240
    Total303060

    12/30=2/5.

    2/5.
  4. Find P(girl|cycle) from the table.
    2 marks
    Invented survey of 60 students; every student is in exactly one travel category.
    TravelGirlBoyTotal
    Cycle12820
    Walk182240
    Total303060

    12/20=3/5.

    3/5.
  5. Using the travel table, explain why P(cycle|girl) and P(girl|cycle) have different denominators.
    3 marks
    Invented survey of 60 students; every student is in exactly one travel category.
    TravelGirlBoyTotal
    Cycle12820
    Walk182240
    Total303060

    Girls total 30, cyclists total 20.

    The same overlap 12 is divided by different totals.

    They condition on different groups.
  6. P(A and B)=0.12 and P(B)=0.3. Find P(A|B).
    2 marks

    0.12/0.3=0.4.

    0.4.
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Curriculum reference & lesson scope

Australian Curriculum v 9 / SA Curriculum

Compound events; systematic counting and the multiplication principle; factorials and simple arrangements; conditional language and restricted sample spaces; conditional probabilities from tables/trees; replacement and dependence; conditional versus unconditional probabilities; designing and interpreting simulations.

Official curriculum source ↗
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