Solutions — Probability Rules

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1. Q1 — Addition rule

   (a) P(A∪B) = 0.3+0.5−0.1 = 0.7   (b) P(A∪B) = 1/4+1/3−1/12 = 3/12+4/12−1/12 = 6/12 = 1/2   (c) P(A∪B) = 0.6+0.7−0.4 = 0.9   (d) Mutually exclusive: P(A∪B) = 0.25+0.35 = 0.60

2. Q2 — Complement rule

   (a) P(A′) = 1−0.72 = 0.28   (b) P(A) = 1−3/7 = 4/7   (c) P(rain) = 1−0.35 = 0.65   (d) P(fail) = 1−4/5 = 1/5

3. Q3 — Two-way table: Year 11/12 × Science/Arts

   Year 11 total = 18+22 = 40. Year 12 total = 24+16 = 40. Science total = 42. Arts total = 38. Grand total = 80.   (a) P(Year 11) = 40/80 = 1/2   (b) P(Science) = 42/80 = 21/40   (c) P(Year 12 ∩ Arts) = 16/80 = 1/5   (d) P(Year 11 ∪ Science) = 40/80 + 42/80 − 18/80 = 64/80 = 4/5

4. Q4 — Mutually exclusive or not?

   (a) Club and diamond: mutually exclusive (no card is both). P = 13/52+13/52 = 26/52 = 1/2   (b) Heart and queen: NOT mutually exclusive (queen of hearts is both). P = 13/52+4/52−1/52 = 16/52 = 4/13   (c) Prime {2,3,5} and even {2,4,6}: NOT mutually exclusive (2 is in both). P = 3/6+3/6−1/6 = 5/6

5. Q5 — Complete the two-way table

   Male/Employed = 35. Male total = 55 → Male/Unemployed = 55−35 = 20. Female total = 100−55 = 45. Female/Unemployed = 18 → Female/Employed = 45−18 = 27. Employed total = 35+27 = 62. Unemployed total = 20+18 = 38.   (a) Table: M/Emp=35, M/Unemp=20, M total=55; F/Emp=27, F/Unemp=18, F total=45; Emp total=62, Unemp total=38, Grand=100.   (b) P(Employed) = 62/100 = 0.62   (c) P(Female ∩ Employed) = 27/100 = 0.27   (d) P(Male ∪ Employed) = 55/100+62/100−35/100 = 82/100 = 0.82

6. Q6 — Finding unknown probability

   (a) P(A∩B) = P(A)+P(B)−P(A∪B) = 0.50+0.40−0.75 = 0.15   (b) P(B) = P(A∪B)−P(A)+P(A∩B) = 0.82−0.60+0.24 = 0.46   (c) Mutually exclusive: P(A)+P(B) = 0.9. P(A)=3P(B): 3P(B)+P(B)=0.9 → 4P(B)=0.9 → P(B)=0.225, P(A)=0.675

7. Q7 — Multi-step: Maths and English

   (a) 10 study neither → 40 study at least one. P(Maths∪English) = 40/50 = 0.8   (b) P(Maths∩English) = P(Maths)+P(English)−P(Maths∪English) = 0.6+0.7−0.8 = 0.5   (c) n(both) = 0.5×50 = 25 students   (d) Maths only = n(Maths) − n(both) = 30−25 = 5 students

8. Q8 — Partition rule: factory machines

   (a) P(Defective) = P(Defective|A)P(A) + P(Defective|B)P(B) = 0.05×0.6 + 0.08×0.4 = 0.03+0.032 = 0.062   (b) P(Not defective) = 1−0.062 = 0.938   (c) P(Machine A ∩ Defective) = 0.05×0.6 = 0.03   (d) Not mutually exclusive. A product can be from Machine A AND be defective simultaneously. P(A∩Defective) = 0.03 ≠ 0.

9. Q9 — Two-way table: age × smartphone

   U30/Smartphone=75, U30/No phone=120−75=45, U30 total=120. 30+/Smartphone=48, 30+/No phone=80−48=32, 30+ total=80. Smartphone total=123, No phone total=77, Grand=200.   (a) Table complete as above.   (b) P(Smartphone) = 123/200 = 0.615   (c) P(U30∪Smartphone) = 120/200+123/200−75/200 = 168/200 = 0.84   (d) P(30+∩No smartphone) = 32/200 = 0.16

10. Q10 — Multi-event reasoning

    (a) Exhaustive & mutually exclusive: 2k+3k+4k=1 → 9k=1 → k=1/9. P(A)=2/9, P(B)=3/9=1/3, P(C)=4/9.   (b) P(A∩B) = 0.5+0.4−0.7 = 0.2. De Morgan: P(A′∪B′) = P((A∩B)′) = 1−0.2 = 0.8.   (c) A only = A∩B′; B only = B∩A′. Since A = (A∩B)∪(A∩B′) and they are mutually exclusive: P(A∩B′)=P(A)−P(A∩B). Similarly P(B∩A′)=P(B)−P(A∩B). Sum = P(A)+P(B)−P(A∩B) = P(A∪B) ✓.