T3T2 Review Solutions — Probability

1. Basic probability. Fluency

   • (a) P(red) = 5/10 = 1/2.
   • (b) P(not blue) = 7/10.
   • (c) (5 + 2)/10 = 7/10.
   • (d) Yes — a single marble cannot be both red and blue. P(R ∩ B) = 0.

2. Addition rule. Fluency

   • (a) P(F) = 18/30 = 3/5.
   • (b) P(F ∪ S) = 18/30 + 12/30 − 6/30 = 24/30 = 4/5.
   • (c) P(neither) = 1 − 4/5 = 1/5.
   • (d) F only = 18 − 6 = 12. P = 12/30 = 2/5.

3. Conditional probability. Fluency

   • (a) P(A|B) = 0.15/0.5 = 0.3.
   • (b) P(A)×P(B) = 0.4×0.6 = 0.24 = P(A∩B). Yes, independent.
   • (c) Red cards = 26. Aces that are red = 2. P(Ace|red) = 2/26 = 1/13.
   • (d) Aces = 4. Red aces = 2. P(red|Ace) = 2/4 = 1/2.

4. Probability distributions. Fluency

   • (a) 0.3 + 0.5 + k = 1 ⇒ k = 0.2.
   • (b) E(X) = 1(0.3) + 2(0.5) + 3(0.2) = 0.3 + 1.0 + 0.6 = 1.9.
   • (c) 1/3 + 1/2 + k = 1 ⇒ 5/6 + k = 1 ⇒ k = 1/6.
   • (d) P(double) = 6/36 = 1/6. E(X) = 1×(1/6) + 0×(5/6) = 1/6 ≈ 0.167.

5. Two-way table — exercise habits. Understanding

   • (a) P(E ∩ D) = 90/200 = 9/20 = 0.45.
   • (b) P(D|E) = 90/120 = 3/4 = 0.75.
   • (c) P(E|D) = 90/120 = 3/4 = 0.75. (Diet total = 90 + 30 = 120.)
   • (d) P(E)×P(D) = (120/200)×(120/200) = 0.6×0.6 = 0.36. P(E∩D) = 0.45 ≠ 0.36. Not independent — exercising and healthy diet are associated.

6. Drawing without replacement. Understanding

   • (a) P(YY) = (4/10)(3/9) = 12/90 = 2/15.
   • (b) P(YP) = (4/10)(6/9) = 24/90 = 4/15.
   • (c) P(YP) + P(PY) = 24/90 + 24/90 = 48/90 = 8/15.
   • (d) P(at least one P) = 1 − P(YY) = 1 − 2/15 = 13/15.

7. Tree diagram — two-stage experiment. Understanding

   • (a) P(pass both) = 0.7×0.6 = 0.42.
   • (b) P(pass M only) + P(pass S only) = 0.7×0.4 + 0.3×0.6 = 0.28 + 0.18 = 0.46.
   • (c) P(fail both) = 0.3×0.4 = 0.12.
   • (d) Since the tests are independent, knowing the student passed maths gives no information about science. P(pass science | pass maths) = 0.6.

8. Expected value in context. Understanding

   • (a) 0.20 + 0.35 + 0.30 + 0.10 + 0.05 = 1.00. ✓ All P ≥ 0.
   • (b) E(X) = 0(0.20) + 1(0.35) + 2(0.30) + 3(0.10) + 4(0.05) = 0 + 0.35 + 0.60 + 0.30 + 0.20 = 1.45 goals/game.
   • (c) P(X ≥ 2) = 0.30 + 0.10 + 0.05 = 0.45.
   • (d) Expected total goals = 20 × 1.45 = 29 goals.

9. Bayes’ theorem — testing for defects. Problem Solving

   • (a) P(D) = 0.05. P(T|D) = 0.90. P(T|D′) = 0.05.
   • (b) P(T) = 0.05×0.90 + 0.95×0.05 = 0.045 + 0.0475 = 0.0925.
   • (c) P(D|T) = (0.05×0.90)/0.0925 = 0.045/0.0925 ≈ 0.486 ≈ 48.6%.
   • (d) A positive test result means there is approximately a 48.6% chance the item is defective. Despite the 90% accuracy, the equal false-positive rate means roughly half of positive tests are false alarms — because most items (95%) are not defective to begin with.

10. Expected value — game design. Problem Solving

    • (a) P($500) = 1/1000. P($100) = 2/1000. P($20) = 5/1000. P($0) = 992/1000.
    • (b) E(X) = 500(1/1000) + 100(2/1000) + 20(5/1000) = 0.50 + 0.20 + 0.10 = $0.80.
    • (c) Total revenue = 1000×$2 = $2000. Total prizes = $500+$200+$100 = $800. Total profit = $1200. (E(profit per ticket) = $2−$0.80 = $1.20.)
    • (d) Required: price − E(prize) = $1.20 ⇒ price = $1.20 + $0.80 = $2.00. The current $2 ticket price is already the correct price.

11. Multi-step conditional probability. Problem Solving

    • (a) P(D∩A) = 0.50×0.02 = 0.010. P(D∩B) = 0.30×0.03 = 0.009. P(D∩C) = 0.20×0.05 = 0.010.
    • (b) P(D) = 0.010 + 0.009 + 0.010 = 0.029.
    • (c) P(C|D) = 0.010/0.029 ≈ 0.345 ≈ 34.5%.
    • (d) P(A|D) = 0.010/0.029 ≈ 34.5%. P(B|D) = 0.009/0.029 ≈ 31.0%. P(C|D) ≈ 34.5%. Machines A and C are equally likely to have produced a defective item.

12. Distribution meets conditional probability (Monty Hall). Problem Solving

    • (a) P(X=1) = 1/3. Only 1 of 3 doors has the car.
    • (b) E(X) = 1×(1/3) + 0×(2/3) = 1/3.
    • (c) Case 1: contestant picked car door (P=1/3). Host opens either goat door. Switching → loses. Case 2: contestant picked goat door 1 (P=1/3). Host must open the other goat door. Switching → wins. Case 3: contestant picked goat door 2 (P=1/3). Host opens the other goat door. Switching → wins. P(win by switching) = 2/3. ✓
    • (d) E(cars won) = 90 × 2/3 = 60 cars.