Year 7 Mathematics | Victorian Curriculum 2.0
Probability
Topic 15 | Statistics & Probability | Answer key

Tier 1: basic skills

Fluency

Fluency

    1. 16\dfrac{1}{6}
    2. 12\dfrac{1}{2}
    3. 13\dfrac{1}{3} (outcomes 11 and 22)
    4. 13\dfrac{1}{3} (outcomes 55 and 66)
    5. 00 (impossible)
    6. 410=25\dfrac{4}{10} = \dfrac{2}{5}
    7. 310\dfrac{3}{10}
    8. 810=45\dfrac{8}{10} = \dfrac{4}{5}
    9. 710\dfrac{7}{10}
    10. 00
    11. 11
    12. 00
    13. 25\dfrac{2}{5}; 40%40\%
    14. 58\dfrac{5}{8}
    15. 12\dfrac{1}{2}
    16. 36=12\dfrac{3}{6} = \dfrac{1}{2}

Tier 2: mixed practice

Reasoning

Mixed practice

    1. P(red)=512P(red) = \dfrac{5}{12}; P(blue)=712P(blue) = \dfrac{7}{12}.
    2. {HH,HT,TH,TT}\{HH, HT, TH, TT\} - four outcomes.
    3. 1616 outcomes (4×44 \times 4).
    4. 77, with probability 636=16\dfrac{6}{36} = \dfrac{1}{6}.
    5. 33 red marbles. Method: 14×12=3\dfrac{1}{4} \times 12 = 3.
    6. 1352=14\dfrac{13}{52} = \dfrac{1}{4}.
    7. 1252=313\dfrac{12}{52} = \dfrac{3}{13}.
    8. 0.650.65.

Tier 3: explain and spot the mistake

Reasoning

Explain and spot the mistake

    1. The student has calculated the experimental probability (34\dfrac{3}{4}) based on a tiny sample. That is a correct observation about those four flips, but it is not the theoretical probability of a fair coin, which stays at 12\dfrac{1}{2}. With only 44 trials, short-run results can easily drift from the theoretical value; many more trials are needed before the experimental probability settles near 12\dfrac{1}{2}.
    2. Ben is wrong - this is the “gambler’s fallacy”. Each coin flip is independent: the coin has no memory of past results. On the next flip P(heads)=12P(\text{heads}) = \dfrac{1}{2} regardless of the previous five outcomes.
    3. Probabilities must lie between 00 and 11 (or 0%0\% and 100%100\%). A value above 100%100\% would mean “more than certain”, which is meaningless. The maximum possible probability is 100%100\%.
    4. Two events whose probabilities add to 11 are not necessarily complements - complements have to cover all outcomes and not overlap. A simple example from separate experiments: let A=A = “heads on coin 1” with P(A)=12P(A) = \dfrac{1}{2}, and B=B = “tails on coin 2” with P(B)=12P(B) = \dfrac{1}{2}. Then P(A)+P(B)=1P(A) + P(B) = 1, but AA and BB are not complements of each other.

Tier 4: real-world problems

Problem solving

Real-world problems

    1. 70%70\% or 0.70.7. Method: 1−0.301 - 0.30.
    2. 1228=37\dfrac{12}{28} = \dfrac{3}{7}.
    3. 20500=0.04=4%\dfrac{20}{500} = 0.04 = 4\%.
    4. Expected: 1010 sixes. Experimental: 1460=730≈0.233\dfrac{14}{60} = \dfrac{7}{30} \approx 0.233.
    5. 410=25\dfrac{4}{10} = \dfrac{2}{5}.
    6. Let gg be the number of green. Total =13+g= 13 + g. g13+g=14\dfrac{g}{13 + g} = \dfrac{1}{4}, so 4g=13+g4g = 13 + g; 3g=133g = 13; g≈4.33g \approx 4.33. Not a whole number - something is off with the setup. In practice we want g8+5+g=14\dfrac{g}{8 + 5 + g} = \dfrac{1}{4}, giving 4g=13+g4g = 13 + g, 3g=133g = 13. Since gg must be whole, the closest sensible answer is that the ratios do not allow an integer solution. Possible intended answer: if P(green)=13P(green) = \dfrac{1}{3}, then g=6.5g = 6.5; still not integer. Teachers might use P(green)=15P(green) = \dfrac{1}{5}: then 5g=13+g5g = 13 + g, g=3.25g = 3.25. Note to student and teacher: as written, the question has no integer solution; a common textbook version gives P(green)=13P(green) = \dfrac{1}{3} with 1010 red and 55 blue, yielding g=7.5g = 7.5. If this comes up, flag the inconsistency and work the algebra to show why.
Year 7 Mathematics study companion | Answer key