/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Two construction contracts are t... [FREE SOLUTION] | 91Ó°ÊÓ

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Two construction contracts are to be randomly assigned to one or more of three firms: \(\mathrm{I}, \mathrm{II},\) and III. Any firm may receive both contracts. If each contract will yield a profit of \(\$ 90,000\) for the firm, find the expected profit for firm I. If firms I and II are actually owned by the same individual, what is the owner's expected total profit?

Short Answer

Expert verified
The expected profit for Firm I is $60,000; for the owner of both firms I and II, it's $120,000.

Step by step solution

01

Define Possible Outcomes

There are two contracts, and each can be assigned to any of the three firms. Therefore, each contract has 3 options for assignment. Let's denote the contracts as A and B. Possible outcomes for assignment are: (I, I), (I, II), (I, III), (II, I), (II, II), (II, III), (III, I), (III, II), and (III, III). These represent which firms contract A and B are assigned to, respectively.
02

Calculate Probabilities

Since the contracts are assigned randomly, each of the 9 possible outcomes is equally likely. Thus, the probability of each outcome is \[ P = \frac{1}{9} \]
03

Determine Profit for Firm I

For each outcome, determine if Firm I gets one or both contracts: - (I, I): Firm I gets both contracts. Profit = 2 x $90,000. - (I, II), (I, III), (II, I), (III, I): Firm I gets one contract in each. Profit = $90,000. - (II, II), (II, III), (III, II), (III, III): Firm I gets no contracts. Profit = 0.
04

Calculate Expected Profit for Firm I

The expected profit \(E\) for Firm I is calculated by summing the profit for each outcome times its probability:\[ E = 1/9 \times 180,000 \ + \ 1/9 \times 90,000 \ + \ 1/9 \times 90,000 \ + \ 1/9 \times 90,000 \ + \ 1/9 \times 90,000 \ + 1/9 \times 0 \ + 1/9 \times 0 \ + 1/9 \times 0 \ + 1/9 \times 0 = 60,000 \]
05

Evaluate Outcomes for Firm I and II Owned Together

If Firm I and Firm II are owned by the same individual, consider the outcomes where either firm I or II receive(s) the contract(s): - (I, I), (I, II), (I, III), (II, I): Owner's total profit is $180,000. - (II, II), (II, III), (III, I), (III, II): Owner's total profit is $90,000. - (III, III): Owner's total profit is 0.
06

Calculate Expected Total Profit for Owner

The expected total profit \(E\) for the owner is:\[ E = 1/9 \times 180,000 + 1/9 \times 180,000 + 1/9 \times 180,000 + 1/9 \times 180,000 + 1/9 \times 90,000 + 1/9 \times 90,000 + 1/9 \times 90,000 + 1/9 \times 90,000 + 1/9 \times 0 = 120,000 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The expected value in probability and statistics is a vital concept. It serves as the average outcome one can anticipate over a series of events. Think of it as a predicted win or loss when playing a game of chance many times over. To calculate the expected value, you multiply each outcome by its probability, and then sum these products. This gives a weighted average of all possible outcomes. In the example of the construction contracts, the firm can expect a certain profit based on how the contracts are distributed. For Firm I, the expected profit is calculated by considering what profit Firm I would have in each possible scenario of contract assignments. Because each contract scenario has an equal probability of occurring, which is \( \frac{1}{9} \) for each of the nine possibilities, the expected profit is calculated by adding up the products of each potential profit outcome and its probability, resulting in: \[ E = \frac{1}{9} \times 180,000 + 4 \times \frac{1}{9} \times 90,000 + 4 \times \frac{1}{9} \times 0 = 60,000 \] This simplifies to an average profit of $60,000 for Firm I.
Random Assignment
Random assignment is a method used to designate subjects into different groups, ensuring that each subject has an equal chance of assignment to any given group or condition. It's a critical component for maintaining fairness and objectivity in experiments and scenarios. In the case of assigning the construction contracts, random assignment means each firm has an equal chance of receiving a contract. This results in nine equally probable outcomes, as each of the three firms can potentially receive either or both of the two contracts. Random assignment ensures no firm is unfairly advantaged or disadvantaged. Having this randomness helps establish the foundations for probability calculations, such as expected value, because it means that each scenario is fair and equally likely, simplifying the calculation to just averaging out evenly across all possible outcomes.
Profit Calculation
Profit calculation is a straightforward yet essential process in business. It's the process of determining the earnings after deducting costs and expenditures. Here, we focus specifically on calculating profits based on outcomes of assigned contracts.Each contract brings in a defined profit of $90,000 for a firm. To find the total profit, you identify how many contracts a firm receives under each scenario and multiply the number by the profit from a single contract. For example, if Firm I receives both contracts (scenario (I, I)), its total profit would be \( 2 \times 90,000 = 180,000 \). If Firm I receives one contract (such as scenario (I, II)), then the profit is simply \( 90,000 \).However, if Firm I receives no contracts in a certain scenario, the profit is zero. By understanding each possible outcome's profits, businesses can make informed decisions and expectations about average profits via techniques like expected value.

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Most popular questions from this chapter

This exercise demonstrates that, in general, the results provided by Tchebysheff's theorem cannot be improved upon. Let \(Y\) be a random variable such that $$p(-1)=\frac{1}{18}, \quad p(0)=\frac{16}{18}, \quad p(1)=\frac{1}{18}$$ a. Show that \(E(Y)=0\) and \(V(Y)=1 / 9\) b. Use the probability distribution of \(Y\) to calculate \(P(|Y-\mu| \geq 3 \sigma) .\) Compare this exact probability with the upper bound provided by Tchebysheff's theorem to see that the bound provided by Tchebysheff's theorem is actually attained when \(k=3\) *.c. In part.(b) we guaranteed \(E(Y)=0\) by placing all probability mass on the values \(-1,0,\) and1, with \(p(-1)=p(1) .\) The variance was controlled by the probabilities assigned to \(p(-1)\) and \(p(1) .\) Using this same basic idea, construct a probability distribution for a random variable \(X\) that will yield \(P\left(\left|X-\mu_{X}\right| \geq 2 \sigma_{X}\right)=1 / 4\) * d. If any \(k>1\) is specified, how can a random variable \(W\) be constructed so that \(P\left(\left|W-\mu_{W}\right| \geq k \sigma_{W}\right)=1 / k^{2} ?\)

Find the distributions of the random variables that have each of the following moment-generating functions: a. \(m(t)=\left[(1 / 3) e^{t}+(2 / 3)\right]^{5}\) b. \(m(t)=\frac{e^{t}}{2-e^{t}}\) c. \(m(t)=e^{2\left(e^{t}-1\right)}\)

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