L32 Solutions — Conditional Probability and Independence

1. Conditional probability basics. Fluency

   • (a) P(A|B) = 0.12/0.4 = 0.3.
   • (b) P(A ∩ B) = P(B)×P(A|B) = 0.6×0.3 = 0.18.
   • (c) Greater than 2: {3,4,5,6} — 4 outcomes. Even and >2: {4,6} — 2 outcomes. P = 2/4 = 1/2.
   • (d) Face cards: J, Q, K = 12. Kings = 4. P(King|face) = 4/12 = 1/3.

2. Testing independence. Fluency

   • (a) P(A)×P(B) = 0.3×0.5 = 0.15 = P(A∩B). Yes, independent.
   • (b) P(A)×P(B) = 0.4×0.6 = 0.24 ≠ 0.28. Not independent.
   • (c) Yes — the outcome of one coin has no effect on the other.
   • (d) No — after the first card is drawn, the deck has 51 cards, so probabilities for the second draw change.

3. With and without replacement. Fluency

   • (a) (3/10)×(3/10) = 9/100 = 0.09.
   • (b) (3/10)×(2/9) = 6/90 = 1/15 ≈ 0.067.
   • (c) Without replacement, the first red ball is not returned, leaving fewer red balls and fewer total balls, so the second probability is lower.
   • (d) P(RB) + P(BR) = (5/10)(5/9) + (5/10)(5/9) = 25/90 + 25/90 = 50/90 = 5/9.

4. Two-way table conditional probabilities. Fluency

   • (a) 48/60 = 4/5 = 0.8.
   • (b) 48/72 = 2/3 ≈ 0.667.
   • (c) P(Exercise)=60/120=0.5. P(Diet)=72/120=0.6. P(E)×P(D)=0.3. P(E∩D)=48/120=0.4 ≠ 0.3. Not independent.
   • (d) 36/60 = 3/5 = 0.6.

5. Tree diagram (4R 6W, without replacement). Understanding

   • (a) 12/90 + 24/90 + 24/90 + 30/90 = 90/90 = 1. ✓
   • (b) P(2nd red) = P(RR) + P(WR) = 12/90 + 24/90 = 36/90 = 2/5.
   • (c) P(1st R | 2nd R) = P(RR)/P(2nd R) = (12/90)/(36/90) = 12/36 = 1/3.
   • (d) Not independent. P(2nd red | 1st red) = 3/9 = 1/3, but P(2nd red) = 2/5. Since 1/3 ≠ 2/5, the draws are dependent.

6. Dependent events. Understanding

   • (a) P(brown eyes | left-handed) = P(both)/P(left-handed) = (4/30)/(8/30) = 4/8 = 1/2.
   • (b) After 1 defective taken: 4 defective left from 19. P = 4/19.
   • (c) P(A∩B) = 0.6×0.4 = 0.24. P(A′∩B) = 0.4×0.2 = 0.08.
   • (d) P(B) = P(A∩B) + P(A′∩B) = 0.24 + 0.08 = 0.32.

7. Independence in context. Understanding

   • (a) Yes — the coin and die are physically separate; neither outcome affects the other.
   • (b) P(rain)×P(traffic) = 0.3×0.5 = 0.15 ≠ 0.2. Not independent — rain increases the probability of a traffic jam.
   • (c) (4/10)×(6/10) = 24/100 = 0.24.
   • (d) (4/10)×(6/9) = 24/90 = 4/15 ≈ 0.267.

8. Conditional from Venn (Car and Transport). Understanding

   • (a) P(C|T) = P(C∩T)/P(T) = (60/300)/(120/300) = 60/120 = 1/2.
   • (b) P(T|C) = 60/180 = 1/3.
   • (c) P(C)×P(T) = (180/300)×(120/300) = 0.6×0.4 = 0.24. P(C∩T) = 60/300 = 0.2. 0.24 ≠ 0.2. Not independent.
   • (d) C only = 120, T only = 60, both = 60. Total at least one = 240. Neither = 300−240 = 60. P = 60/300 = 1/5.

9. Medical testing (2% prevalence). Problem Solving

   • (a) P(D) = 0.02. P(+|D) = 0.95. P(+|D′) = 0.08.
   • (b) P(+) = 0.02×0.95 + 0.98×0.08 = 0.019 + 0.0784 = 0.0974.
   • (c) P(D|+) = (0.02×0.95)/0.0974 = 0.019/0.0974 ≈ 0.195 ≈ 19.5%.
   • (d) Even with a positive test result, there is only about a 19.5% chance of actually having the disease. This surprises people because the test seems accurate, but when the disease is rare, most positive results are false positives.

10. Quality control (two machines). Problem Solving

    • (a) P(def ∩ A) = 0.60×0.03 = 0.018.
    • (b) P(def ∩ B) = 0.40×0.05 = 0.020.
    • (c) P(def) = 0.018 + 0.020 = 0.038.
    • (d) P(A | def) = 0.018/0.038 = 18/38 = 9/19 ≈ 0.474 ≈ 47.4%.