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A telemarketer is planning to set up a phone bank to bilk widows with a Ponzi scheme. His past experience (prior to his most recent incarceration) suggests that each phone will be in use half the time. For a given phone at a given time, let 0 indicate that the phone is available and let 1 indicate that a caller is on the line. Suppose that the telemarketer's "bank" is comprised of four telephones. (a) Write out the outcomes in the sample space. (b) What outcomes would make up the event that exactly two phones are being used? (c) Suppose the telemarketer had \(k\) phones. How many outcomes would allow for the possibility that at most one more call could be received? (Hint: How many lines would have to be busy?)

Step by step solution

01

Identify the Binary Outcomes

For a single phone, there are two possible outcomes: the phone being in use (1) or the phone being available (0).

02

Compute the Sample Space

For a bank of four phones, the sample space consists of the different combinations of these phones being in use or not. Since each phone can be in two states, and there are four phones, there would be \(2^4 = 16\) possible outcomes. These would be: 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111.

03

Determine Outcomes for Two Phones Being Used

To get exactly two phones being used, we want combinations of two 1s and two 0s. These would be: 0011, 0101, 0110, 1001, 1010, 1100. So, there are 6 outcomes.

04

Determine Outcomes for At Most One More Call

If a maximum of one more call could be received, it means that all but one of the phones are busy (1). For a telemarketer with \(k\) phones, there are \(k\) ways this can happen (one for each phone that could possibly be available).

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Key Concepts

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

Sample Space

Probability theory often involves the concept of a sample space, which is the complete set of all possible outcomes of a probabilistic scenario. In the context of the telemarketer's phone bank, each phone can either be busy or available. Since there are four phones, the total number of configurations can be represented as all combinations of phones being busy (represented by '1') or available (represented by '0').

To compute the sample space, we consider all possible states for the four phones:

• Each phone can exist in one of two states: busy (1) or available (0).
• The total number of outcomes for four phones is calculated by multiplying the number of states each phone can be in, which is mathematically expressed as: \(2^4 = 16\).
• The resulting sample space includes the outcomes: 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111.

Grasping the concept of a sample space is crucial as it forms the foundation for identifying probabilities of different events in probability theory.

Binary Outcomes

In this scenario, each phone's state can be represented by binary outcomes, a fundamental aspect of probability theory and combinatory mathematics. Binary outcomes mean there are exactly two possible states or results:

• A phone being in use, denoted by '1'.
• A phone being available, denoted by '0'.

Binary outcomes make it easier to model and calculate probabilities because they simplify complex events into basic building blocks. Each "0" or "1" represents a distinct phone's state at a given moment, allowing us to create a simple yet powerful framework for analyzing the event's probability. For example:

• To find scenarios where exactly two phones are in use, consider the patterns like: 0011, 0101, 0110, 1001, 1010, 1100.
• These represent all configurations where exactly two "1" outcomes occur in the string.

Binary outcomes are key in determining possibilities in situations involving "]two state" variables like switches, coins, or indeed phones.

Combinatory Mathematics

Combinatory mathematics provides us with tools to count outcomes and efficiently handle complex probability scenarios. In this exercise, combinatory techniques help find how many ways certain configurations can occur without listing each possibility.

For instance, to determine how many ways at most one phone can take another call, we harness combinatorics:

• If there are \(k\) phones and at most one more call can be received, it implies all other phones are busy.
• This is equivalent to choosing one phone to be available, which is a simple selection of 1 phone from \(k\), and thus \(k\) configurations.

When trying to find how many outcomes exist for exactly two busy phones, combinatorics can also help:

• We calculate the number of combinations of taking 2 busy phones out of 4, denoted as \(\binom{4}{2}\).
• The calculation yields 6, matching our previous listing of outcomes: 0011, 0101, 0110, 1001, 1010, 1100.

Understanding combinatory mathematics is essential for simplifying and solving complex probability problems without exhaustive lists.