L33 Solutions — Probability Distributions

1. Valid distributions. Fluency

   • (a) 0.2 + 0.5 + 0.3 = 1.0; all values between 0 and 1. Yes, valid.
   • (b) 0.3 + 0.4 + k = 1 ⇒ k = 0.3.
   • (c) 1/8 + 3/8 + 3/8 + k = 1 ⇒ 7/8 + k = 1 ⇒ k = 1/8.
   • (d) No. Probabilities must satisfy P ≥ 0. A negative probability has no meaning.

2. Expected value. Fluency

   • (a) E(X) = 0(0.4) + 1(0.4) + 2(0.2) = 0 + 0.4 + 0.4 = 0.8.
   • (b) E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7.
   • (c) E(X) = (1+2+3+4+5+6)/6 = 21/6 = 3.5.
   • (d) E(X) = −1(1/4) + 0(1/2) + 1(1/4) = −1/4 + 0 + 1/4 = 0.

3. Game expectation. Fluency

   • (a) E(prize) = 10(0.05) + 0(0.95) = $0.50. Net = $0.50 − $1.00 = −$0.50 per ticket. The buyer loses 50c on average.
   • (b) Cost = $2. P(win) = 1/500. E(net) = 400(1/500) − 2 = 0.80 − 2 = −$1.20.
   • (c) E(net) = 3(1/2) + (−2)(1/2) = 1.5 − 1 = +$0.50. The player expects to gain 50c per flip.
   • (d) E(net winnings) = 0. A fair game gives neither side any long-run advantage.

4. Distributions from scenarios. Fluency

   • (a) S = {HH, HT, TH, TT}. P(X=0) = 1/4, P(X=1) = 2/4 = 1/2, P(X=2) = 1/4.
   • (b) P(X=1) = 1/5.
   • (c) Odd outcomes (1,3,5) all give X=0; P(X=0)=1/2. Even outcomes give X=2,4,6 each with P=1/6. Distribution: P(0)=1/2, P(2)=1/6, P(4)=1/6, P(6)=1/6.
   • (d) E(X) = 0(1/2) + 2(1/6) + 4(1/6) + 6(1/6) = 0 + 2/6 + 4/6 + 6/6 = 12/6 = 2.

5. Probability distribution from a diagram. Understanding

   • (a) 0.15 + 0.35 + 0.30 + 0.15 + 0.05 = 1.00. ✓
   • (b) P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) = 0.30 + 0.15 + 0.05 = 0.50.
   • (c) E(X) = 0(0.15) + 1(0.35) + 2(0.30) + 3(0.15) + 4(0.05) = 0 + 0.35 + 0.60 + 0.45 + 0.20 = 1.60 pets.
   • (d) The most likely value is X = 1. This is called the mode.

6. Distribution from counting. Understanding

   • (a) X = 0, 1, or 2.
   • (b) P(X=0) = (2/5)(1/4) = 2/20 = 1/10. P(X=1) = (3/5)(2/4) + (2/5)(3/4) = 6/20 + 6/20 = 3/5. P(X=2) = (3/5)(2/4) = 6/20 = 3/10.
   • (c) 1/10 + 6/10 + 3/10 = 10/10 = 1. ✓ E(X) = 0(1/10) + 1(3/5) + 2(3/10) = 0 + 6/10 + 6/10 = 6/5 = 1.2.
   • (d) On average, you would expect to draw 1.2 red balls per pair of draws from this box.

7. Expected winnings. Understanding

   • (a) P($0) = 6/12 = 1/2. P($5) = 3/12 = 1/4. P($10) = 2/12 = 1/6. P($50) = 1/12.
   • (b) E(X) = 0(1/2) + 5(1/4) + 10(1/6) + 50(1/12) = 0 + 15/12 + 20/12 + 50/12 = 85/12 ≈ $7.08.
   • (c) E(operator profit per spin) = $4 − E(prize) = $4 − $7.08 = −$3.08. The operator loses $3.08 per spin on average — this game favours the player.
   • (d) Since the operator has a negative expected profit (−$3.08/spin), no number of spins yields expected profit for the operator. The prize pool would need to be reduced or the cost increased.

8. Comparing distributions. Understanding

   • (a) E(A) = 200(0.5) + 50(0.3) + (−100)(0.2) = 100 + 15 − 20 = $95.
   • (b) E(B) = 120(0.7) + (−50)(0.3) = 84 − 15 = $69.
   • (c) Option A has the higher expected gain ($95 vs $69).
   • (d) A risk-averse investor should choose Option B — it has a smaller spread of outcomes and avoids the possibility of a $100 loss, even though its expected gain is lower.

9. Designing a fair game. Problem Solving

   • (a) P(sum=7): {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6/36 = 1/6. P(sum=11): {(5,6),(6,5)} = 2/36 = 1/18.
   • (b) P(win) = 6/36 + 2/36 = 8/36 = 2/9.
   • (c) E(net) = 10(2/9) + 0(7/9) − c = 20/9 − c.
   • (d) Fair when E(net) = 0: 20/9 − c = 0 ⇒ c = 20/9 ≈ $2.22.

10. Insurance expected value. Problem Solving

    • (a) E(payout) = 150000(0.002) + 3000(0.05) + 0(0.948) = 300 + 150 = $450.
    • (b) E(profit) = $800 − $450 = $350 per policy.
    • (c) Total profit = 350 × 10 000 = $3 500 000.
    • (d) Insurance protects against a catastrophic, unaffordable loss. While the expected value for the homeowner is negative (−$350/yr), the alternative — a $150 000 loss with no coverage — could be financially devastating. People pay a small certain cost to avoid a small chance of ruin.