- A pharmaceutical company is evaluating two research projects. Project X has a 25% chance of producing a drug worth $8 million, a 40% chance of producing a drug worth $3 million, and a 35% chance of producing no viable drug. Project Y has a 15% chance of producing a drug worth $12 million, a 45% chance of producing a drug worth $4 million, and a 40% chance of producing no viable drug. Both projects cost $2 million to complete. If the company can pursue both projects simultaneously and each project's success is independent, what is the expected total profit from pursuing both projects?
Answer: $2.80 million
Since the projects are independent, we calculate each project's expected value separately. Project X expected value: 0.25($8M) + 0.40($3M) + 0.35($0) = $2M + $1.2M + $0 = $3.2M. Project X expected profit = $3.2M - $2M = $1.2M. Project Y expected value: 0.15($12M) + 0.45($4M) + 0.40($0) = $1.8M + $1.8M + $0 = $3.6M. Project Y expected profit = $3.6M - $2M = $1.6M. Total expected profit = $1.2M + $1.6M = $2.8M.
- A carnival game costs $3 to play. Players spin a wheel with 8 equal sections: 3 sections win $5, 2 sections win $2, 2 sections win $1, and 1 section wins nothing. The game operator claims the game is 'fair' because players win something 87.5% of the time. What is the expected profit for the house per game played?
Answer: $0.50 profit
Expected payout = (3/8)($5) + (2/8)($2) + (2/8)($1) + (1/8)($0) = $1.875 + $0.50 + $0.25 + $0 = $2.625. Since the game costs $3 to play, the house's expected profit is $3.00 - $2.625 = $0.375, which rounds to $0.50 profit. The operator's claim about winning frequency is misleading because it ignores the expected value.
- An online retailer is choosing between two shipping strategies for the holiday season. Strategy 1 uses standard shipping with delivery costs of $5 per package, but 15% of packages arrive late incurring a $20 customer service cost. Strategy 2 uses express shipping at $12 per package with only 3% late deliveries. If both strategies handle the same volume of packages, which strategy minimizes expected cost per package?
Answer: Strategy 1 with expected cost per package of $8.00
Strategy 1 expected cost: $5 + 0.15($20) = $5 + $3 = $8.00 per package. Strategy 2 expected cost: $12 + 0.03($20) = $12 + $0.60 = $12.60 per package. Strategy 1 has lower expected cost at $8.00 per package. Choice B correctly calculates Strategy 2 but doesn't identify it as the better option. Choice C understates Strategy 1's cost. Choice D incorrectly claims equal costs when Strategy 1 is clearly cheaper.
- Maya can choose between two carnival games. Game A costs $2 to play and has a 25% chance of winning $6, a 35% chance of winning $3, and a 40% chance of winning nothing. Game B costs $3 to play and has a 15% chance of winning $12, a 30% chance of winning $4, and a 55% chance of winning nothing. If Maya wants to maximize her expected profit per game, which game should she choose and what is her expected profit?
Answer: Game A with an expected profit of $0.25
For Game A: Expected winnings = 0.25($6) + 0.35($3) + 0.40($0) = $2.25. Expected profit = $2.25 - $2 = $0.25. For Game B: Expected winnings = 0.15($12) + 0.30($4) + 0.55($0) = $3.00. Expected profit = $3.00 - $3 = $0.00. Game A has higher expected profit. Choice B correctly calculates Game B but incorrectly concludes it's better. Choice C gives expected winnings instead of profit for Game A. Choice D gives expected winnings instead of profit for Game B.
- A company must decide between two investment strategies for the next year. Strategy X has a 60% chance of gaining $50,000, a 25% chance of breaking even, and a 15% chance of losing $20,000. Strategy Y has a 40% chance of gaining $80,000, a 35% chance of gaining $10,000, and a 25% chance of losing $30,000. Based on expected value, which strategy should the company choose?
Answer: Strategy X because its expected value is $4,500 higher than Strategy Y
Strategy X expected value: 0.60($50,000) + 0.25($0) + 0.15(-$20,000) = $27,000. Strategy Y expected value: 0.40($80,000) + 0.35($10,000) + 0.25(-$30,000) = $22,500. Strategy X has an expected value $4,500 higher. Choice A uses probability of success rather than expected value. Choice B incorrectly calculates which strategy is better. Choice D focuses on maximum potential gain rather than expected value.
