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Chapter 2: Problem 2

A part selected for testing is equally likely to have been produced on any one of six cutting tools. a. What is the sample space? b. What is the probability that the part is from tool \(1 ?\) c. What is the probability that the part is from tool 3 or tool \(5 ?\) d. What is the probability that the part is not from tool 4 ?

Short Answer

Expert verified
a) \( S = \{1, 2, 3, 4, 5, 6\} \) b) \( \frac{1}{6} \) c) \( \frac{1}{3} \) d) \( \frac{5}{6} \).

Step by step solution

01

Define the Sample Space

The sample space is the set of all possible outcomes. Since the part can be produced by any one of the six tools, the sample space is: \( S = \{1, 2, 3, 4, 5, 6\} \).
02

Calculate Probability for Tool 1

Each tool is equally likely to produce the part, so each tool has a probability of \( \frac{1}{6} \). The probability that the part is from tool 1 is \( \frac{1}{6} \).
03

Calculate Probability for Tool 3 or Tool 5

To find the probability that the part is from either tool 3 or tool 5, add the probabilities for each tool. Since the probability for each tool is \( \frac{1}{6} \), then probability(3 or 5) = \( \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \) which simplifies to \( \frac{1}{3} \).
04

Calculate Probability for Not Tool 4

Find the probability that the part is not from tool 4 by considering all tools except tool 4. There are 5 such outcomes: \( \{1, 2, 3, 5, 6\} \).The probability is, therefore, \( \frac{5}{6} \).

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Key Concepts

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

Understanding the Sample Space
In probability theory, the concept of a "sample space" is fundamental. It refers to the set of all possible outcomes of an experiment. When you perform an experiment or a test, there could be multiple outcomes. Collectively, these possibilities are known as the sample space.
In the provided exercise, we are testing parts produced on six different cutting tools. Each tool represents a unique outcome. Therefore, the sample space for this specific situation is represented as:
  • Tool 1
  • Tool 2
  • Tool 3
  • Tool 4
  • Tool 5
  • Tool 6
Thus, the sample space is the set \( S = \{1, 2, 3, 4, 5, 6\} \). Each number corresponds to a particular cutting tool from which a part can potentially originate.
Equally Likely Outcomes
When we say outcomes are "equally likely," it means that each outcome has the same chance of occurring. This concept is significant because it simplifies the calculation of probabilities.
In the exercise, each part is equally likely to be produced by any of the six tools. Since there are no preferences or biases towards any single tool, we say they are equally likely.
  • There are six possible outcomes (or tools).
  • Each tool has a probability of \( \frac{1}{6} \) of producing the part.
This equal likelihood allows us to directly use \( \frac{1}{6} \) as the probability for each tool when calculating specific probabilities such as parts from tool 1 or any other single tool.
Calculating Outcome Probability
Outcome probability is the likelihood of a specific event occurring. In this case, it鈥檚 finding the probability that a part is produced by a certain tool.
Let's look at two scenarios from the exercise:
  • **Probability of Part from Tool 1:** Since each tool has an equal chance, the probability that a part is from tool 1 is \( \frac{1}{6} \).

  • **Probability of Part from Tool 3 or Tool 5:** To find the probability for tool 3 or 5, we add their individual probabilities. Both these events, tool 3 and tool 5, are independent, and adding them gives us a combined probability of \( \frac{1}{3} \) since \( \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \), which simplifies to \( \frac{1}{3} \).
Calculating probabilities this way provides a simple method to determine the likelihood of one or more specific events.
Exploring the Complement Rule
The complement rule in probability is a handy tool for finding the probability that an event does not occur, using the event that it does occur. It鈥檚 based on the principle that the total probability of all possible outcomes in a sample space is 1.
If the probability of an event occurring is \( P(A) \), the probability of it not occurring (its complement) is \( 1 - P(A) \).
In the exercise, we apply this rule to find the probability that a part is not from tool 4:
  • The probability for tool 4 is \( \frac{1}{6} \).
  • Therefore, the probability that a part is not from tool 4 is \( 1 - \frac{1}{6} = \frac{5}{6} \).
This shows the power of the complement rule in simplifying probability calculations by focusing on what doesn't happen, rather than what does.

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Most popular questions from this chapter

In the manufacturing of a chemical adhesive, \(3 \%\) of all batches have raw materials from two different lots. This occurs when holding tanks are replenished and the remaining portion of a lot is insufficient to fill the tanks. Only \(5 \%\) of batches with material from a single lot require reprocessing. However, the viscosity of batches consisting of two or more lots of material is more difficult to control, and \(40 \%\) of such batches require additional processing to achieve the required viscosity. Let \(A\) denote the event that a batch is formed from two different lots, and let \(B\) denote the event that a lot requires additional processing. Determine the following probabilities: a. \(P(A)\) b. \(P\left(A^{\prime}\right)\) c. \(P(B \mid A)\) d. \(P\left(B \mid A^{\prime}\right)\) e. \(P(A \cap B)\) g. \(P(B)\)

A batch of 500 containers of frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement, from the batch. Let \(A\) and \(B\) denote the events that the first and second containers selected are defective, respectively. a. Are \(A\) and \(B\) independent events? b. If the sampling were done with replacement, would \(A\) and \(B\) be independent?

Semiconductor lasers used in optical storage products require higher power levels for write operations than for read operations. High-power-level operations lower the useful life of the laser. Lasers in products used for backup of higher-speed magnetic disks primarily write, and the probability that the useful life exceeds five years is \(0.95 .\) Lasers that are in products that are used for main storage spend approximately an equal amount of time reading and writing, and the probability that the useful life exceeds five years is \(0.995 .\) Now, \(25 \%\) of the products from a manufacturer are used for backup and \(75 \%\) of the products are used for main storage. Let \(A\) denote the event that a laser's useful life exceeds five years, and let \(B\) denote the event that a laser is in a product that is used for backup. Use a tree diagram to determine the following: a. \(P(B)\) b. \(P(A \mid B)\) c. \(P\left(A \mid B^{\prime}\right)\) d. \(P(A \cap B)\) e. \(P\left(A \cap B^{\prime}\right)\) f. \(P(A)\) g. What is the probability that the useful life of a laser exceeds five years? h. What is the probability that a laser that failed before five years came from a product used for backup?

Computer keyboard failures are due to faulty electrical connects (12\%) or mechanical defects (88\%). Mechanical defects are related to loose keys \((27 \%)\) or improper assembly \((73 \%)\) Electrical connect defects are caused by defective wires (35\%), improper connections \((13 \%),\) or poorly welded wires \((52 \%)\) a. Find the probability that a failure is due to loose keys. b. Find the probability that a failure is due to improperly connected or poorly welded wires.

The edge roughness of slit paper products increases as knife blades wear. Only \(1 \%\) of products slit with new blades have rough edges, \(3 \%\) of products slit with blades of average sharpness exhibit roughness, and \(5 \%\) of products slit with worn blades exhibit roughness. If \(25 \%\) of the blades in manufacturing are new, \(60 \%\) are of average sharpness, and \(15 \%\) are worn, what is the proportion of products that exhibit edge roughness?
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