Finite Mathematics Quiz: Decision Trees And Rollback
8 questions · exam conditions
0:00
Decision Trees And RollbackQuestion 1 of 8

A company is considering launching a new product. The company can either launch immediately or first conduct a market survey for a cost of $10,000. If they launch immediately, analysts estimate a 50% chance of high demand (profit of $180,000) and a 50% chance of low demand (loss of $60,000). If they conduct the survey, there is a 60% chance of a positive result and a 40% chance of a negative result. After a positive result, the company can choose to launch, with an 80% chance of high demand (profit of $200,000) and a 20% chance of low demand (loss of $50,000). After a negative result, the company can still choose to launch, with a 30% chance of high demand (profit of $150,000) and a 70% chance of low demand (loss of $80,000). In all cases where the company has the option not to launch, the payoff for not launching is $0.

Based on a rollback analysis using expected monetary value (EMV), what is the optimal initial decision and the overall expected value of the project?

Conduct the survey; the project's EMV is $90,000.
Conduct the survey; the project's EMV is $80,000.
Do not conduct the survey; the project's EMV is $60,000.
Do not conduct the survey; the project's EMV is $120,000.
← Back to quizzes

Finite Mathematics Quiz

Finite Mathematics Quiz: Decision Trees And Rollback

Practice Decision Trees And Rollback in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Decision Trees And Rollback, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A company is considering launching a new product. The company can either launch immediately or first conduct a market survey for a cost of $10,000. If they launch immediately, analysts estimate a 50% chance of high demand (profit of $180,000) and a 50% chance of low demand (loss of $60,000). If they conduct the survey, there is a 60% chance of a positive result and a 40% chance of a negative result. After a positive result, the company can choose to launch, with an 80% chance of high demand (profit of $200,000) and a 20% chance of low demand (loss of $50,000). After a negative result, the company can still choose to launch, with a 30% chance of high demand (profit of $150,000) and a 70% chance of low demand (loss of $80,000). In all cases where the company has the option not to launch, the payoff for not launching is $0.

Based on a rollback analysis using expected monetary value (EMV), what is the optimal initial decision and the overall expected value of the project?

  1. Conduct the survey; the project's EMV is $90,000.
  2. Conduct the survey; the project's EMV is $80,000. (correct answer)
  3. Do not conduct the survey; the project's EMV is $60,000.
  4. Do not conduct the survey; the project's EMV is $120,000.
Explanation: The problem is solved using rollback analysis. First, evaluate the 'Launch Immediately' branch: EMV(Launch Immediately) = 0.5(\180,000) + 0.5(-$60,000) = $90,000 - $30,000 = $60,000.Since$60,000>$0(fornotlaunching),thevalueofthispathis$60,000.Next,evaluatethe′ConductSurvey′branch.Ifthesurveyispositive,EMV(Launch)=. Since $60,000 > $0 (for not launching), the value of this path is $60,000. Next, evaluate the 'Conduct Survey' branch. If the survey is positive, EMV(Launch) = 0.8($200,000) + 0.2(-$50,000) = $160,000 - $10,000 = $150,000.Thisisgreaterthan$0,sothedecisionistolaunch.Ifthesurveyisnegative,EMV(Launch)=. This is greater than $0, so the decision is to launch. If the survey is negative, EMV(Launch) = 0.3($150,000) + 0.7(-$80,000) = $45,000 - $56,000 = -$11,000.Thisislessthan$0,sothedecisionisnottolaunch(valueis$0).Theexpectedvalueatthesurvey′schancenodeis. This is less than $0, so the decision is not to launch (value is $0). The expected value at the survey's chance node is 0.6($150,000) + 0.4($0) = $90,000.Finally,subtractthesurveycost:. Finally, subtract the survey cost: $90,000 - $10,000 = $80,000$. Comparing the initial options, EMV(Survey) = $80,000 is greater than EMV(Launch Immediately) = $60,000. Thus, the optimal decision is to conduct the survey, and the project's EMV is $80,000.

Question 2

A student is deciding on a study strategy for a final exam. The student can either 'Cram' the night before or join a 'Study Group'. The outcomes are measured in utility points (utils). If the student chooses to Cram, there is a 60% chance it is effective (resulting in a Grade A, 100 utils) and a 40% chance it is ineffective (Grade C, 50 utils). If the student chooses the Study Group, there is a 50% chance the group is focused (Grade A, 100 utils) and a 50% chance it is distracted (Grade B, 80 utils).

To maximize expected utility, which strategy should the student choose, and what is the expected utility of that choice?

  1. Study Group; Expected Utility = 90 (correct answer)
  2. Cram; Expected Utility = 80
  3. Study Group; Expected Utility = 92
  4. Cram; Expected Utility = 75
Explanation: The expected utility (EU) for each strategy is calculated by summing the products of each outcome's utility and its probability. For 'Cram': EU(Cram) = 0.6(100)+0.4(50)=60+20=800.6(100) + 0.4(50) = 60 + 20 = 80 utils. For 'Study Group': EU(Study Group) = 0.5(100)+0.5(80)=50+40=900.5(100) + 0.5(80) = 50 + 40 = 90 utils. Comparing the two expected utilities, EU(Study Group) = 90 is greater than EU(Cram) = 80. Therefore, the optimal strategy is to join the Study Group, which has an expected utility of 90.

Question 3

A farmer must decide whether to plant Corn or Soybeans. Planting Corn has a cost of $10,000, and planting Soybeans has a cost of $15,000. The revenue for each crop depends on whether the coming season is wet or dry. The probability of a wet season is 0.4, and the probability of a dry season is 0.6. The projected revenues are as follows: Corn yields $50,000 in a wet season and $80,000 in a dry season. Soybeans yield $70,000 in a wet season and $60,000 in a dry season.

Based on the expected monetary value, which crop should the farmer plant, and what is the expected profit?

  1. Plant Corn; Expected Profit = $68,000
  2. Plant Soybeans; Expected Profit = $64,000
  3. Plant Soybeans; Expected Profit = $49,000
  4. Plant Corn; Expected Profit = $58,000 (correct answer)
Explanation: First, calculate the expected profit (EMV) for each option by considering revenues and costs. For Corn: The profit in a wet season is \50,000 - $10,000 = $40,000.Theprofitinadryseasonis. The profit in a dry season is $80,000 - $10,000 = $70,000.EMV(Corn)=. EMV(Corn) = 0.4($40,000) + 0.6($70,000) = $16,000 + $42,000 = $58,000.ForSoybeans:Theprofitinawetseasonis. For Soybeans: The profit in a wet season is $70,000 - $15,000 = $55,000.Theprofitinadryseasonis. The profit in a dry season is $60,000 - $15,000 = $45,000.EMV(Soybeans)=. EMV(Soybeans) = 0.4($55,000) + 0.6($45,000) = $22,000 + $27,000 = $49,000$. Comparing the two, EMV(Corn) = $58,000 is greater than EMV(Soybeans) = $49,000. Therefore, the farmer should plant Corn for an expected profit of $58,000.

Question 4

A software company is deciding between developing an app in-house or outsourcing it. In-house development costs $150,000 and has a 70% chance of producing a high-quality app and a 30% chance of a low-quality app. Outsourcing costs $100,000 and there is an 80% chance the contractor is reliable and a 20% chance they are unreliable. A reliable contractor always produces a high-quality app. An unreliable contractor has a 40% chance of producing a high-quality app and a 60% chance of a low-quality one. A high-quality app generates $300,000 in revenue, while a low-quality app generates $50,000.

What is the optimal decision based on expected monetary value, and what is that value?

  1. In-house; EMV = $75,000
  2. Outsource; EMV = $190,000
  3. Outsource; EMV = $170,000 (correct answer)
  4. Outsource; EMV = $250,000
Explanation: First, calculate net payoffs (Revenue - Cost). In-house: HighQ profit = $300k - $150k = $150k; LowQ profit = $50k - 150k=−150k = -100k. EMV(In-house) = $0.7(150k) + 0.3(-100k) = 105k - 30k = $75k. Outsource: HighQ profit = $300k - $100k = $200k; LowQ profit = $50k - 100k=−100k = -50k. The outsource path has a two-stage chance process. If unreliable, the EMV of the outcome is $0.4(200k) + 0.6(-50k) = 80k - 30k = $50k. Now, roll back the first chance node for the outsource option: EMV(Outsource) = $P(Reliable) \times (Profit if reliable\text{Profit if reliable}) + P(Unreliable) \times (EMV if unreliable\text{EMV if unreliable})$. EMV(Outsource) = $0.8(200k) + 0.2(50k) = 160k + 10k = $170k. Comparing the two options, EMV(Outsource) = $170k is greater than EMV(In-house) = $75k. The optimal decision is to outsource.

Question 5

A company must decide between building a small plant or a large plant. The profitability of each depends on future market demand, which can be high or low, with equal probability (P(High) = 0.5, P(Low) = 0.5). A small plant yields a profit of $100,000 with high demand and $40,000 with low demand. A large plant yields a profit of $250,000 with high demand but a loss of $50,000 with low demand.

A consulting firm offers a new technology that would only affect the profit of the small plant in a high-demand market. What is the minimum profit the small plant with high demand would need to generate for the optimal decision to switch from 'Build Large' to 'Build Small'?

  1. $100,000
  2. $120,000
  3. $160,000 (correct answer)
  4. $200,000
Explanation: First, calculate the current EMV for each option. EMV(Small) = 0.5(\100,000) + 0.5($40,000) = $50,000 + $20,000 = $70,000.EMV(Large)=. EMV(Large) = 0.5($250,000) + 0.5(-$50,000) = $125,000 - $25,000 = $100,000.Currently,theoptimaldecisionistobuildthelargeplant.ThedecisionwillswitchwhentheEMVofthesmallplantisequaltoorgreaterthantheEMVofthelargeplant.Let. Currently, the optimal decision is to build the large plant. The decision will switch when the EMV of the small plant is equal to or greater than the EMV of the large plant. Let X be the new profit for the small plant with high demand. We set the new EMV(Small) equal to EMV(Large): $$0.5(X) + 0.5(\40,000) = $100,000 0.5X + $20,000 = $100,000 0.5X = $80,000 X = $160,000$$ Therefore, the profit for the small plant with high demand must be at least $160,000.

Question 6

A simplified decision tree analysis for a project resulted in an overall expected monetary value of $150. The initial decision is between Alternative A, which has a certain payoff of $140, and Alternative B. Alternative B leads to a chance node with a 50% probability of a 'High' outcome (payoff $250) and a 50% probability of a 'Low' outcome (payoff $X).

Given that the overall EMV of the project is $150, what must be the value of the payoff $X$?

  1. $30
  2. $50 (correct answer)
  3. $70
  4. $150
Explanation: The overall EMV of the project is the value at the initial decision node. This value is the maximum of the values of the branches, so EMV(Project) = max(Value(A), Value(B)). We are given Value(A) = $140 and EMV(Project) = $150. Since $150 > $140, the optimal decision must have been to choose Alternative B, and the value of Alternative B must be $150. The value of Alternative B is its expected monetary value: EMV(B) = 0.5(\250) + 0.5(X).Wesetthisequalto$150:$. We set this equal to $150: $$150 = 0.5($250) + 0.5X $150 = $125 + 0.5X $25 = 0.5X X = $50

Question 7

An investor is considering two projects, Project X and Project Y. A decision tree analysis has been performed. The optimal strategy is to choose Project X, which has an expected monetary value (EMV) of $50,000. The EMV for Project Y is $40,000. The analysis for Project X is based on a probability of success $p=0.6$. The investor wants to know how sensitive the optimal decision is to this probability estimate.

Let the payoff for success in Project X be 100,000andthepayoffforfailurebe−100,000 and the payoff for failure be -25,000. The decision to choose Project X over Project Y remains optimal as long as the probability of success pp is greater than or equal to what threshold value?

  1. p=0.520p = 0.520 (correct answer)
  2. p=0.400p = 0.400
  3. p=0.600p = 0.600
  4. p=0.680p = 0.680
Explanation: When you encounter decision tree sensitivity analysis problems, you're testing how changes in probability estimates affect the optimal choice between alternatives. The key is finding the probability threshold where both projects have equal expected value. To find this threshold, you need to set up an equation where Project X's EMV equals Project Y's EMV. Project X's expected value is: p×$100,000+(1−p)×(−$25,000)p \times \$100,000 + (1-p) \times (-\$25,000). This simplifies to $125,000p−$25,000\$125,000p - \$25,000. Project Y has a fixed EMV of $40,000. Setting them equal: $\125,000p - $25,000 = $40,000 . Solving for p: $125,000p = $65,000 , so p = 0.520 . This means Project X remains optimal when p \geq 0.520 . Choice A (p=0.520p = 0.520) is correct—this is the exact breakeven point where both projects have equal expected value. Choice B (p=0.400p = 0.400) is wrong because this represents a probability below the threshold. At this level, Project Y would actually be superior to Project X. Choice C (p=0.600p = 0.600) is wrong because this is the original probability given in the problem, not the threshold value we're solving for. Choice D (p=0.680p = 0.680) is wrong because this exceeds the threshold unnecessarily. While Project X would still be optimal at this probability, it's not the minimum required threshold. Study tip: In sensitivity analysis problems, always set up equations where competing alternatives have equal expected values to find decision thresholds. The threshold represents the point where you're indifferent between choices.

Question 8

In a standard decision tree used for rollback analysis, which of the following actions is performed at a square decision node?

  1. The node's value is calculated by taking a weighted average of the values of the subsequent nodes using their probabilities.
  2. The cost of reaching the node is subtracted from the simple average of the values of the subsequent nodes.
  3. The node's value is determined by the outcome with the highest probability among all branches originating from it.
  4. The node's value is determined by selecting the maximum value among the immediate subsequent nodes. (correct answer)
Explanation: Decision tree analysis is a fundamental tool in decision-making under uncertainty, where you work backward from outcomes to determine optimal choices. The key distinction is between square nodes (decision points where you choose) and circular nodes (chance events with given probabilities). At square decision nodes, you have control over which path to take, so you naturally want to choose the option that gives you the best possible outcome. This means selecting the branch with the maximum expected value among all immediate options available from that node. You're literally picking the best choice available to you at that moment. Option A describes what happens at circular chance nodes, not square decision nodes. At chance nodes, you calculate expected value using weighted averages based on probabilities since you don't control the outcome. Option B is incorrect because decision analysis doesn't involve subtracting costs from simple averages of subsequent values—this misunderstands both the cost accounting and averaging processes. Option C confuses probability with value maximization; even if an outcome has the highest probability, it might not have the highest expected value, and probability-based selection isn't how decision nodes work anyway. The correct answer is D because decision nodes represent points where you actively choose the best available option, which means selecting the maximum value path. Study tip: Remember the shape-function relationship: squares = decisions = maximum selection, circles = chance = probability-weighted averaging. When you see a square node, always think "choose the best option available."