Topic Review — Probability — Solutions

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1. Tree diagrams and multi-step probability

   1. P(both red, with replacement): P(R)=3/5. P(RR) = 3/5 × 3/5 = 9/25
   2. P(at least one Head, 2 coins): 1 − P(TT) = 1 − 1/4 = 3/4
   3. P(both green, without replacement): 4/5 × 3/4 = 12/20 = 3/5
   4. Sample space size: 3 × 4 = 12 outcomes

2. Venn diagrams

   1. Venn and P(neither): M only=12, T only=6, both=6, neither=30−12−6−6=6. P(neither)=6/30=1/5
   2. P(M ∪ T): 24/30 = 4/5
   3. P(A ∪ B): 0.5+0.4−0.2 = 0.7
   4. Mutually exclusive?: No — P(A ∩ B) = 0.2 ≠ 0.

3. Two-way table — music and instruments

   1. P(Rock AND plays): 18/80 = 9/40
   2. P(plays | Rock): 18/28 = 9/14
   3. P(Pop OR plays): P(Pop)+P(plays)−P(Pop∩plays) = 36/80+40/80−14/80 = 62/80 = 31/40
   4. Independence check: P(plays)×P(Rock)=40/80×28/80=1120/6400=0.175. P(plays∩Rock)=18/80=0.225. 0.175≠0.225, so not independent.

4. Experimental probability and bias

   1. Expected frequency of 3: 1/6 × 180 = 30
   2. Experimental P(3): 42/180 = 7/30 ≈ 0.233
   3. Comparison: Theoretical = 1/6 ≈ 0.167. Experimental = 0.233. The die rolled 3 about 40% more often than expected. This suggests possible bias, though more trials are needed.
   4. Why 1800 is better: By the law of large numbers, with more trials the experimental probability converges to the theoretical one (if the die is fair). With 1800 rolls, random fluctuation contributes less relative error, making any genuine bias much more detectable.

5. Mixed problem solving

   1. Language Venn:
      1. F only=7, J only=5, both=3, neither=10. Check: 7+5+3+10=25 ✓
      2. P(F∪J) = (7+3+5)/25 = 15/25 = 3/5
      3. P(J|F) = P(J∩F)/P(F) = (3/25)/(10/25) = 3/10
   2. James and rain:
      1. Tree: Rain(0.6) → Lift: Late(0.05), Not Late(0.95). No Rain(0.4) → Walk: Late(0.3), Not Late(0.7).
      2. P(late) = P(Rain)×P(late|Rain) + P(No Rain)×P(late|Walk) = 0.6×0.05 + 0.4×0.3 = 0.03+0.12 = 0.15
      3. P(Rain|late) = P(Rain∩late)/P(late) = 0.03/0.15 = 1/5 = 0.2
   3. Spinner fairness: Theoretical: each colour 500/3 ≈ 167. Experimental: Red=180/500=0.36, Blue=165/500=0.33, Green=155/500=0.31. These are reasonably close to 1/3≈0.333. The student’s claim has merit — with 500 trials, this level of variation is normal. The spinner is not clearly biased, but more data would be needed for certainty.