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Five firms, \(F_{1}, F_{2}, \ldots, F_{5},\) each offer bids on three separate contracts, \(C_{1}, C_{2},\) and \(C_{3} .\) Any one firm will be awarded at most one contract. The contracts are quite different, so an assignment of \(C_{1}\) to \(F_{1},\) say, is to be distinguished from an assignment of \(C_{2}\) to \(F_{1}\). a. How many sample points are there altogether in this experiment involving assignment of contracts to the firms? (No need to list them all.) b. Under the assumption of equally likely sample points, find the probability that \(F_{3}\) is awarded a contract.

Short Answer

Expert verified
a: 60 sample points. b: Probability is \(\frac{3}{5}\).

Step by step solution

01

Determine Possible Assignments for Contract C1

For the first contract, \(C_1\), there are 5 firms \(F_1, F_2, F_3, F_4, F_5\) available to choose from. Therefore, there are 5 possible choices for \(C_1\).
02

Determine Possible Assignments for Contract C2

Once a firm is assigned to \(C_1\), only 4 firms are left available for \(C_2\). Thus, there are 4 possible choices for assigning \(C_2\).
03

Determine Possible Assignments for Contract C3

After assigning firms to \(C_1\) and \(C_2\), only 3 firms remain for the third contract \(C_3\). Therefore, \(C_3\) has 3 choices available.
04

Compute Total Number of Sample Points

The total number of ways to assign the contracts to the firms can be calculated by multiplying the number of choices at each step: \(5 \times 4 \times 3 = 60\). Therefore, there are 60 sample points in the experiment.
05

Determine Probability of Firm F3 Receiving a Contract

Each firm, including \(F_3\), can receive only one of three contracts. If \(F_3\) is awarded any of the contracts, calculate the number of favorable outcomes: - Assign \(C_1\) to \(F_3\): 4 remaining firms for \(C_2\), then 3 for \(C_3\) - giving \(4 \times 3 = 12\) possibilities. - Assign \(C_2\) to \(F_3\): 4 firms for \(C_1\) and 3 for \(C_3\) - giving \(4 \times 3 = 12\) possibilities.- Assign \(C_3\) to \(F_3\): 4 firms for \(C_1\) and then 3 for \(C_2\) - giving \(4 \times 3 = 12\) possibilities.Thus, the total favorable outcomes for \(F_3\) receiving any contract is \(12 + 12 + 12 = 36\).
06

Calculate the Probability of F3 Being Awarded a Contract

The probability that \(F_3\) is awarded a contract is calculated by dividing the number of favorable outcomes by the total number of sample points: \(\frac{36}{60} = \frac{3}{5}\). So, the probability is \(\frac{3}{5}\).

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Key Concepts

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

Sample Space
In probability theory, a sample space is the set of all possible outcomes or results that can occur in a given experiment. When considering the assignment of contracts to firms, each unique arrangement of assignments forms an individual outcome within the sample space.

To understand this better using our exercise, observe the process of assigning contracts to firms. For the first contract, any of the five firms (\(F_1\) to \(F_5\)) can be selected, each choice forming a possible outcome. With five firms available for the first contract (\(C_1\)), after assigning it, only four firms remain for the second contract (\(C_2\)), and finally three firms for the third contract (\(C_3\)).

The multiplicative principle of counting demonstrates the sample space size, calculated as:
  • 5 choices for \(C_1\)
  • 4 choices for \(C_2\)
  • 3 choices for \(C_3\)
Therefore, the total number of possible combinations or sample points in this experiment is:\[5 \times 4 \times 3 = 60\]

Understanding the sample space helps visualize all potential ways contracts can be assigned, laying a foundation for calculating probabilities.
Combinatorics
Combinatorics is a branch of mathematics concerning the counting, arrangement, and combination of elements within a set. It plays a crucial role in solving problems related to probabilities by determining how many possible arrangements or selections can be made.

In our contract-assignment scenario, combinatorics is applied to figure out how many different ways we can assign the contracts to the firms. By using the principle of permutations, which considers the order of assignments, we multiply the number of choices available at each stage:
  • 5 options for the first contract (\(C_1\))
  • After one firm is chosen, 4 options for the second (\(C_2\))
  • Finally, 3 options for the third (\(C_3\))
Thus, the computation \(5 \times 4 \times 3 = 60\) tells us there are 60 unique ways to assign all the contracts.

Combinatorics simplifies the task of counting scenarios in complex operations, making it easier to determine probabilities when dealing with various assignments or selections.
Conditional Probability
In probability, conditional probability is the measure of the probability of an event occurring given that another event has already occurred. When attempting to find the probability of a firm receiving a contract, we consider each contract assignment conditionally.

Using firm \(F_3\) as an example, we are interested in evaluating the probability that this specific firm secures a contract. Let's break down the steps:
  • If \(F_3\) is awarded \(C_1\), there are 12 ways to assign the other two contracts to the remaining firms.
  • If \(F_3\) gets \(C_2\), similarly, there are 12 configurations for the others \((4 \times 3 = 12)\).
  • If \(F_3\) receives \(C_3\), again 12 configurations emerge.
Adding these, we find 36 favorable outcomes for \(F_3\) attaining any contract.

Thus, the conditional probability that \(F_3\) secures a contract, given the total sample space of 60 possible assignments, is calculated as:\[\frac{36}{60} = \frac{3}{5}\]

This concept helps in understanding how the likelihood of an event changes when additional conditions or information are considered, a crucial concept in statistics and decision-making.

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Most popular questions from this chapter

Articles coming through an inspection line are visually inspected by two successive inspectors. When a defective article comes through the inspection line, the probability that it gets by the first inspector is .1. The second inspector will "miss" five out of ten of the defective items that get past the first inspector. What is the probability that a defective item gets by both inspectors?

An experimenter wishes to investigate the effect of three variables \(-\) pressure, temperature, and the type of catalyst-on the yield in a refining process. If the experimenter intends to use three settings each for temperature and pressure and two types of catalysts, how many experimental runs will have to be conducted if he wishes to run all possible combinations of pressure, temperature, and types of catalysts?

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If \(P(A)>0, P(B)>0,\) and \(P(A)

-Suppose that \(A\) and \(B\) are independent events such that the probability that neither occurs is \(a\) and the probability of \(B\) is \(b\). Show that \(P(A)=\frac{1-b-a}{1-b}\).

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