Year 8 Mathematics | Victorian Curriculum 2.0
Probability: complementary & compound events
Topic 16 | Statistics & Probability | Answer key

Year 8 core - answers

Fluency

Complementary events

    1. 0.70.7
    2. 35%35\%
    3. 710\dfrac{7}{10}
    4. 78\dfrac{7}{8}
    5. 56\dfrac{5}{6}. Method: 1−6361 - \dfrac{6}{36}.
Fluency

Two-stage experiments

    1. 44
    2. 1212
    3. 14\dfrac{1}{4}. Method: 36×36\dfrac{3}{6} \times \dfrac{3}{6}.
    4. 336=112\dfrac{3}{36} = \dfrac{1}{12}. Method: (4,6),(5,5),(6,4)(4,6), (5,5), (6,4).
    5. 1136\dfrac{11}{36}. Method: 1−P(no 6)=1−(56)2=1−25361 - P(\text{no 6}) = 1 - \left(\dfrac{5}{6}\right)^2 = 1 - \dfrac{25}{36}.
Fluency

Venn and two-way tables

    1. 1414.
    2. 1010.
    3. 66. Method: 40−24−10−(neither)=40−24−10+10=?40 - 24 - 10 - (\text{neither}) = 40 - 24 - 10 + 10 = ?. Actually: coffee only ++ tea only ++ both ++ neither =40= 40; 14+10+10+?=4014 + 10 + 10 + ? = 40; neither =6= 6.
    4. 3440=1720=0.85\dfrac{34}{40} = \dfrac{17}{20} = 0.85.
    5. 2440=35=0.60\dfrac{24}{40} = \dfrac{3}{5} = 0.60.
Reasoning

Explain and spot the mistake

    1. Not correct. Complementary events must also cover all possibilities (be exhaustive) and not overlap. Counter-example: A=A = “roll a 1”, B=B = “roll a 6” on a fair die. P(A)+P(B)=26≠1P(A) + P(B) = \tfrac{2}{6} \ne 1, and even if both halved the probability to sum to 11, they’d need to overlap zero and exhaust the sample space.
    2. Not the same. Mutually exclusive events cannot both happen. Independent events have no influence on each other. Example: “rolling a 6” and “rolling a 1” are mutually exclusive but not independent in a single roll. “Coin lands heads” and “die rolls 6” are independent but not mutually exclusive.
    3. Because the overlap (both coffee and tea) is counted twice when you add P(coffee)+P(tea)P(\text{coffee}) + P(\text{tea}). You subtract P(both)P(\text{both}) to correct.
    4. No - the gambler’s fallacy. Each flip is independent; the coin has no memory. P(tails next)=0.5P(\text{tails next}) = 0.5.
Problem solving

Real contexts

    1. Pairs: AB, AC, AD, BC, BD, CD - six total. P(Anna)=36=12P(\text{Anna}) = \dfrac{3}{6} = \dfrac{1}{2}.
    2. P(both red)=58×47=2056=514P(\text{both red}) = \dfrac{5}{8} \times \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}.
    3. P(neither)=1−P(sport or instrument)=1−(0.60+0.40−0.25)=1−0.75=0.25P(\text{neither}) = 1 - P(\text{sport or instrument}) = 1 - (0.60 + 0.40 - 0.25) = 1 - 0.75 = 0.25.
    4. P=12×12×16=124P = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{24}.

Challenge - answers

Reasoning

Harder reasoning

    1. 38\dfrac{3}{8}. Method: 33 outcomes with exactly 22 heads out of 88 total (HHT, HTH, THH).
    2. 625\dfrac{6}{25}. Method: P(sum≥8)P(\text{sum} \geq 8): count pairs (3,5),(4,4),(4,5),(5,3),(5,4),(5,5) - six pairs.
    3. P(boy or plays cricket)=0.48+0.30−0.20=0.58P(\text{boy or plays cricket}) = 0.48 + 0.30 - 0.20 = 0.58; P(neither)=1−0.58=0.42P(\text{neither}) = 1 - 0.58 = 0.42.
    4. 836=29\dfrac{8}{36} = \dfrac{2}{9}. Method: pairs with difference 22: (1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4)(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4) - eight.
Year 8 Mathematics study companion | Answer key