Solutions — Conditional Probability and Independence

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1. Q1 — Calculating conditional probability

   (a) P(A|B) = P(A∩B)/P(B) = 0.12/0.4 = 0.3   (b) P(B|A) = P(A∩B)/P(A) = (1/6)/(1/2) = 1/3   (c) P(A∩B) = P(A|B)×P(B) = 0.3×0.5 = 0.15   (d) P(B) = P(A∩B)/P(A|B) = 0.18/0.6 = 0.3

2. Q2 — Conditional probability from two-way table

   Male total=50, Female total=50, Passed total=66, Failed total=34.   (a) P(Passed|Male) = 36/50 = 18/25   (b) P(Male|Passed) = 36/66 = 6/11   (c) P(Failed|Female) = 20/50 = 2/5   (d) P(Female|Failed) = 20/34 = 10/17

3. Q3 — Testing independence

   (a) P(A)×P(B) = 0.4×0.3 = 0.12 = P(A∩B). Independent.   (b) P(A)×P(B) = 0.5×0.6 = 0.30 ≠ 0.25 = P(A∩B). Not independent (dependent).   (c) P(A)×P(B) = (1/3)(1/4) = 1/12 = P(A∩B). Independent.   (d) P(A|B) = 0.5 ≠ 0.7 = P(A). Not independent (dependent).

4. Q4 — Spinner: independence on single spin

   A = {1,3,5}: P(A)=3/5. B = {4,5}: P(B)=2/5. A∩B = {5}: P(A∩B)=1/5. P(A)×P(B)=3/5×2/5=6/25≠1/5=5/25. Not independent.   (b) P(odd on both) = (3/5)×(3/5) = 9/25 (independent spins with replacement)   (c) P(at least one 5) = 1 − P(no 5 on either) = 1−(4/5)² = 1−16/25 = 9/25

5. Q5 — Tree diagram: marbles without replacement

   Bag: 4R, 3B. Total=7.   (a) Branches: RR (4/7×3/6=12/42), RB (4/7×3/6=12/42), BR (3/7×4/6=12/42), BB (3/7×2/6=6/42). Total=42/42 ✓   (b) P(RR) = 12/42 = 2/7   (c) P(exactly one blue) = P(RB)+P(BR) = 12/42+12/42 = 24/42 = 4/7   (d) P(2nd blue|1st red) = 3/6 = 1/2   (e) P(1st red|2nd blue) = P(RB)/P(2nd blue) = (12/42)/(18/42) = 12/18 = 2/3

6. Q6 — Medical testing

   Using 10 000 people: D=100, D′=9900.   Positive∩D = 100×0.95=95. Positive∩D′ = 9900×0.10=990. Total positive=1085.   (a) P(positive) = 1085/10000 = 0.1085   (b) P(D|positive) = 95/1085 ≈ 0.0876 (about 8.8%)   (c) Even though the test correctly identifies 95% of sick people, the disease is rare (only 1% have it). The vast majority of people tested are healthy, and even 10% of the large healthy group gives many false positives. The false positives overwhelm the true positives, making the positive predictive value low. This is the base rate fallacy.

7. Q7 — Independent component failures

   P(A fails)=0.05, P(B fails)=0.08. Independent.   (a) P(both fail) = 0.05×0.08 = 0.004   (b) P(neither fails) = 0.95×0.92 = 0.874. P(machine fails) = 1−0.874 = 0.126   (c) P(only A fails) = P(A)×P(B′) = 0.05×0.92 = 0.046   (d) P(B fails|machine fails) = P(B fails ∩ machine fails)/P(machine fails). P(B fails ∩ machine fails) = P(at least B fails) = 1−P(B works) = ... = P(B alone)+P(both)=0.08×0.95+0.004=0.076+0.004=0.080. P(B|machine fails) = 0.080/0.126 ≈ 0.635

8. Q8 — Three-stage coin flips

   P(H)=0.6, P(T)=0.4.   (a) P(exactly 2H) = P(HHT)+P(HTH)+P(THH) = 3×(0.6×0.6×0.4) = 3×0.144 = 0.432   (b) P(at least 2H) = P(exactly 2H)+P(HHH) = 0.432+0.216 = 0.648   (c) P(3rd H|first two H) = P(H) = 0.6 (flips are independent)   (d) Yes, independent. The coin has no memory; the outcome of one flip does not affect any other. Confirmed by P(3rd H|first two H) = P(H).

9. Q9 — Proving independence from a table

   Total=200. Pass total=120, Male total=100.   (a) P(Pass)=120/200=3/5; P(Male)=100/200=1/2; P(Male∩Pass)=60/200=3/10   (b) P(Male)×P(Pass) = 1/2×3/5 = 3/10 = P(Male∩Pass). Independent. ✓   (c) P(Pass|Male) = 60/100 = 3/5; P(Pass|Female) = 60/100 = 3/5. Both equal 3/5. Knowing the student’s gender provides no information about passing.   (d) Independence means gender has no effect on pass rate. The pass rate is the same (60%) for males and females. Gender and subject outcome are unrelated in this data.

10. Q10 — Extended multiplication rule

    (a) Bag: 5W, 3B. Draw 3 without replacement. P(WWW) = 5/8×4/7×3/6 = 60/336 = 5/28   (b) P(Ace 1st) = 4/52. P(Ace 2nd|Ace 1st) = 3/51. P(Ace 3rd|first two Aces) = 2/50. P(AAA) = 4/52×3/51×2/50 = 24/132600 = 1/5525   (c) P(A∩B∩C) = P(A)×P(B|A)×P(C|A∩B) = 0.4×0.5×0.6 = 0.12