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Counting Four wires (red, green, blue, and yellow) need to be attached to a circuit board. A robotic device will attach the wires. The wires can be attached in any order, and the production manager wishes to determine which order would be fastest for the robot to use. Use the multiplication rule of counting to determine the number of possible sequences of assembly that must be tested. Hint: There are four choices for the first wire, three for the second, two for the third, and only one for the fourth.

Short Answer

Expert verified
There are 24 possible sequences for attaching the wires to the circuit board.

Step by step solution

01

Identify the Choices for the First Wire

When attaching the first wire to the circuit board, the robotic device can choose from any of the four wires: red, green, blue, or yellow. Thus, there are 4 possible choices for the first wire.
02

Determine Choices for the Second Wire

Once the first wire has been attached, there are only three wires remaining that can be selected for attachment. Hence, there are 3 choices for the second wire.
03

Determine Choices for the Third Wire

After attaching two wires, there are only two wires left. Therefore, the robot has 2 choices for the third wire.
04

Attach the Final Wire

Once three wires are attached, there is only one wire left to attach. As a result, there is exactly one choice for the fourth wire.
05

Calculate the Total Number of Sequences

Using the multiplication rule of counting, multiply the number of choices at each step to find the total number of possible sequences: \[ 4 \times 3 \times 2 \times 1 = 24. \]

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Key Concepts

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

Permutations
Permutations are an essential concept in the field of combinatorics. They help us understand the different ways to arrange a set of items where order matters. In the context of attaching wires — Red, Green, Blue, and Yellow — to a circuit board, each unique ordering of wires is a permutation. Since the task involves arranging 4 distinct wires, we determine the number of permutations by considering each possible scenario step by step.

Firstly, you pick from 4 wires for the first position. Once a wire is used, you'll have 3 wires left for the second position, then 2 choices for the third, and finally, only 1 for the last spot. The multiplication rule plays a pivotal role here, enabling us to compute permutations by multiplying the number of choices at each step:
  • 4 choices for the first wire,
  • 3 choices for the second,
  • 2 for the third wire,
  • and 1 for the last wire.
Combining these steps, the total permutations are computed as: \[ 4 \times 3 \times 2 \times 1 = 24 \] possible sequences.
Combinatorics
Combinatorics deals with counting, arrangement, and combination of sets of elements. It's a broad field that helps us solve various counting problems, including permutations and combinations.

In our wire attachment example, the focus is on arrangements where the order is significant, thus forming permutations. Combinatorics provides the tools to efficiently calculate such problems, often using factorials. A factorial, denoted by the symbol \(!\), is the product of all positive integers up to a given number. For 4 wires, the factorial operation is expressed as \[4! = 4 \times 3 \times 2 \times 1\], which results in 24.

This means there are 24 ways to arrange the wires. Such combinatorial approaches are crucial not only in mathematics but also in fields like computer science, logistics, and genetics, where arranging or choosing specific orders and combinations of items is vital.
Probability
Probability is a measure of how likely an event is to occur. When dealing with permutations, it intersects with probability when we assess the chances of one particular sequence occurring among all possible sequences. In the case of attaching wires to a circuit board, knowing each possible sequence allows us to calculate the probability that any given arrangement is chosen.

Given that there are 24 different sequences, the probability of selecting one specific sequence (such as RGBY) randomly is \[ \frac{1}{24} \].

Each sequence is equally likely because the robotic device can place any wire in any order. Understanding probability through permutations is crucial in areas like games, predictions, and statistical modeling, where foreseeing the likelihood of various outcomes is required.

By combining our understanding of permutations and probability, we can effectively arrange items and predict possible outcomes with more accuracy. This kind of analysis is not just academic; it has real-world applications in making decisions and optimizing processes.

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Most popular questions from this chapter

Consider a family with 3 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also \(0.5,\) and that the events "boy" and "girl" are independent. (a) List the equally likely events for the gender of the 3 children, from oldest to youngest. (b) What is the probability that all 3 children are male? Notice that the complement of the event "all three children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female.

Critical Thinking Consider the following events for a driver selected at random from the general population: \(A=\) driver is under 25 years old \(B=\) driver has received a speeding ticket Translate each of the following phrases into symbols. (a) The probability the driver has received a speeding ticket and is under 25 years old (b) The probability a driver who is under 25 years old has received a speeding ticket (c) The probability a driver who has received a speeding ticket is 25 years old or older (d) The probability the driver is under 25 years old or has received a speeding ticket (e) The probability the driver has not received a speeding ticket or is under 25 years old

On a single toss of a fair coin, the probability of heads is 0.5 and the probability of tails is \(0.5 .\) If you toss a coin twice and get heads on the first toss, are you guaranteed to get tails on the second toss? Explain.

Basic Computation: Rules of Probability Given \(P\left(A^{c}\right)=0.8, P(B)=0.3\) \(P(B | A)=0.2:\) (a) Compute \(P(A \text { and } B).\) (b) Compute \(P(A \text { or } B).\)

Probability Estimate: Wiggle Your Ears Can you wiggle your ears? Use the students in your statistics class (or a group of friends) to estimate the percentage of people who can wiggle their ears. How can your result be thought of as an estimate for the probability that a person chosen at random can wiggle his or her ears? Comment: National statistics indicate that about \(13 \%\) of Americans can wiggle their ears (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York).

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