Decision Probability

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Decision Probability · Grade 11 · Worksheet 3

1. Olivia is deciding whether to launch a new eco-friendly product line. If she launches now, there is a 0.35 probability that demand will be high, earning her $45,000 profit, and a 0.65 probability that demand will be low, resulting in a $9,000 loss. If she waits one year, market research will be completed, giving her perfect information about demand. However, waiting will cost $3,000 in lost early-mover advantage, regardless of the outcome. Should Olivia launch now or wait, and what is the expected value of the better decision? Answer: Launch now with expected profit of $9,900 Solution: Calculate expected value if launching now. EV_now = (0.35 * 45000) + (0.65 * (-9000)) = 15750 - 5850 = 9900. So expected profit is $9,900.
   Full step-by-step solution

   Step 1: Calculate expected value if launching now. EV_now = (0.35 * 45000) + (0.65 * (-9000)) = 15750 - 5850 = 9900. So expected profit is $9,900. Step 2: If Olivia waits, she will have perfect information. She will launch only if demand is high (probability 0.35), earning $45,000, and will not launch if demand is low (probability 0.65), earning $0. The expected profit before the waiting cost is (0.35 * 45000) + (0.65 * 0) = 15750. Then subtract the $3,000 cost of waiting: 15750 - 3000 = 12750. Step 3: Compare: EV_now = $9,900, EV_wait = $12,750. Since $12,750 > $9,900, Olivia should wait. Step 4: The expected value of the better decision (waiting) is $12,750. Final answer: Wait, with expected profit of $12,750.

2. Sophia is evaluating two investment strategies. Strategy A: gain $2800 with probability 0.32, lose $600 with probability 0.68. Strategy B: gain $1900 with probability 0.54, lose $400 with probability 0.46. Which strategy has the higher expected value and by how much? Answer: Strategy B by $60 Solution: E(A) = (2800 × 0.32) + (-600 × 0.68) = 896 + (-408) = 488 E(B) = (1900 × 0.54) + (-400 × 0.46) = 1026 + (-184) = 842 E(A) = 488, E(B) = 842 842 - 488 = 354 Since E(B) > E(A), Strategy B has the higher expected value by $354.
   Full step-by-step solution

   Step 1: Calculate expected value for Strategy A E(A) = (2800 × 0.32) + (-600 × 0.68) = 896 + (-408) = 488 Step 2: Calculate expected value for Strategy B E(B) = (1900 × 0.54) + (-400 × 0.46) = 1026 + (-184) = 842 Step 3: Compare the expected values E(A) = 488, E(B) = 842 842 - 488 = 354 Step 4: Determine which strategy is better Since E(B) > E(A), Strategy B has the higher expected value by $354. The answer is Strategy B by $354.

3. Matiu is a farmer deciding between two irrigation systems for his orchard. System A costs $2,400 to install and has a 70% chance of increasing his annual profit by $4,800 and a 30% chance of increasing it by only $1,200 due to variable rainfall. System B costs $1,800 to install and has a 55% chance of increasing his annual profit by $3,600 and a 45% chance of increasing it by $2,400. Based on the expected net profit (profit increase minus installation cost) for one year, which irrigation system should Matiu choose, and what is the expected net profit of that system? Answer: System B with expected net profit of $1,260 Solution: Calculate the expected profit increase for System A. Expected increase = (0.70 * 4800) + (0.30 * 1200) = 3360 + 360 = 3720. Expected net profit = 3720 - 2400 = 1320.
   Full step-by-step solution

   Step 1: Calculate the expected profit increase for System A. Expected increase = (0.70 * 4800) + (0.30 * 1200) = 3360 + 360 = 3720. Expected net profit = 3720 - 2400 = 1320. Step 2: Calculate the expected profit increase for System B. Expected increase = (0.55 * 3600) + (0.45 * 2400) = 1980 + 1080 = 3060. Expected net profit = 3060 - 1800 = 1260. Step 3: Compare expected net profits: System A gives $1,320, System B gives $1,260. Since $1,320 > $1,260, System A has a higher expected net profit. Step 4: The better decision is System A with an expected net profit of $1,320. Final answer: System A with expected net profit of $1,320.

4. Matiu can choose between two investment options. Option A: gain $1200 with probability 0.4, lose $300 with probability 0.6. Option B: gain $800 with probability 0.55, lose $200 with probability 0.45. Which option has the higher expected value and by how much? Answer: Option A by $70 Solution: E(A) = (1200 × 0.4) + (-300 × 0.6) = 480 + (-180) = 300 E(B) = (800 × 0.55) + (-200 × 0.45) = 440 + (-90) = 350 E(A) = 300, E(B) = 350 E(B) - E(A) = 350 - 300 = 50 Option B has the higher expected value by $50 The answer is Option B by $50.
   Full step-by-step solution

   Step 1: Calculate expected value for Option A E(A) = (1200 × 0.4) + (-300 × 0.6) = 480 + (-180) = 300 Step 2: Calculate expected value for Option B E(B) = (800 × 0.55) + (-200 × 0.45) = 440 + (-90) = 350 Step 3: Compare the expected values E(A) = 300, E(B) = 350 E(B) - E(A) = 350 - 300 = 50 Step 4: Determine which is higher and by how much Option B has the higher expected value by $50 The answer is Option B by $50.

5. Mason is organizing a fundraising booth at a school fair and is deciding between two games. In Game A, a player pays $7 to roll a six-sided die. If the die shows a 1 or a 2, the player wins $17; if it shows a 3 or a 4, the player wins $7 (breaking even on the fee); if it shows a 5 or a 6, the player wins nothing. In Game B, a player pays $7 to draw a card from a standard 52-card deck. If the card is a heart, the player wins $12; if it is a club, the player wins $2; if it is a spade or diamond, the player wins nothing. Based on the expected profit for the booth per play (the fee minus the prize), which game should Mason choose to maximize the booth's expected earnings per game? Answer: D. Game B Solution: Calculate expected profit for Game A. Fee = $7. Die outcomes: 1 or 2 (probability 2/6 = 1/3) win $17, so booth profit = 7 - 17 = -$10 (loss).
   Full step-by-step solution

   Step 1: Calculate expected profit for Game A. Fee = $7. Die outcomes: 1 or 2 (probability 2/6 = 1/3) win $17, so booth profit = 7 - 17 = -$10 (loss). 3 or 4 (probability 2/6 = 1/3) win $7, so booth profit = 7 - 7 = $0. 5 or 6 (probability 2/6 = 1/3) win $0, so booth profit = 7 - 0 = $7. Expected profit = (1/3)*(-10) + (1/3)*0 + (1/3)*7 = (-10/3) + 0 + (7/3) = (-3/3) = -$1. So expected loss of $1 per play. Step 2: Calculate expected profit for Game B. Fee = $7. Deck: 52 cards, 13 hearts (probability 13/52 = 1/4) win $12, booth profit = 7 - 12 = -$5. 13 clubs (probability 13/52 = 1/4) win $2, booth profit = 7 - 2 = $5. 26 spades/diamonds (probability 26/52 = 1/2) win $0, booth profit = 7 - 0 = $7. Expected profit = (1/4)*(-5) + (1/4)*5 + (1/2)*7 = (-5/4) + (5/4) + (7/2) = 0 + 3.5 = $3.50. So expected profit of $3.50 per play. Step 3: Compare: Game A expected profit = -$1, Game B expected profit = $3.50. Since $3.50 > -$1, Game B is the better choice. The answer is Game B.