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Chapter 2: Problem 58

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

Understanding the Sample Space

The sample space is the set of all possible outcomes. Since there are six cutting tools, each tool can be represented as an element of the sample space: \( S = \{1, 2, 3, 4, 5, 6\} \).
02

Calculating Probability for Tool 1

Since each tool is equally likely to be selected, the probability that the part is from tool 1 is the ratio of favorable outcomes to the total number of outcomes. Therefore, \( P(\text{Tool 1}) = \frac{1}{6} \).
03

Calculating Probability for Tool 3 or Tool 5

The probability that the part is from tool 3 or tool 5 is the sum of probabilities of the independent events. Each tool has a probability of \( \frac{1}{6} \), so \( P(\text{Tool 3 or Tool 5}) = P(\text{Tool 3}) + P(\text{Tool 5}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \).
04

Calculating Probability of Not Tool 4

To find the probability that the part is not from tool 4, we calculate the probability of the complementary event: \( P(\text{Not Tool 4}) = 1 - P(\text{Tool 4}) = 1 - \frac{1}{6} = \frac{5}{6} \).

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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 denoted as the set of all possible outcomes of a particular random experiment. In the case of our exercise, the sample space refers to the different cutting tools that could have produced the tested part. Since there are a total of six tools, the sample space is represented by the set:
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
Having a clear sample space helps in identifying all potential outcomes, which is crucial for calculating probabilities. In essence, each tool is an element, so the sample space size is 6.
Equally Likely Outcomes
When we talk about equally likely outcomes, it means that each outcome in the sample space has the same probability of occurring. In the exercise, every tool has an equal chance of being selected.
This implies that if you randomly pick a part, the likelihood it comes from any specific tool, say tool 1, is the same as it coming from any other tool.
Mathematically, for our set of six tools, the probability of picking a part from a specific tool, like tool 1, is calculated as the ratio of one favorable outcome to the total number of possible outcomes. This gives us: \[P( ext{Tool 1}) = \frac{1}{6}\]
Complementary Events
Complementary events refer to the outcomes which, together with the event in question, make up the entire sample space. In simpler terms, if you have a probability of something happening, its complement is the probability that it does not happen.
In our exercise, the probability that a part is not from tool 4 is a complementary event to the probability that it is from tool 4. Since there are six tools and one specific tool (tool 4), the complement (not tool 4) means choosing any of the other five tools.
Thus, the probability calculation for not choosing tool 4 is: \[P( ext{Not Tool 4}) = 1 - P( ext{Tool 4}) = 1 - \frac{1}{6} = \frac{5}{6}\]
Independent Events
Independent events are those whose outcomes do not affect each other. In the realm of probability, if the occurrence of one event does not influence the occurrence of another, these events are independent.
In the original problem, when considering the probability that the part comes from either tool 3 or tool 5, these are two separate events. Choosing part from tool 3 is independent of choosing the part from tool 5. Hence, their combined probability is the sum of their individual probabilities:
  • Probability of the part being from tool 3, \( P( ext{Tool 3}) = \frac{1}{6} \)
  • Probability of the part being from tool 5, \( P( ext{Tool 5}) = \frac{1}{6} \)
Thus, the probability of the part being from either tool 3 or tool 5:\[P( ext{Tool 3 or Tool 5}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\]

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Most popular questions from this chapter

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?

A batch of 140 semiconductor chips is inspected by choosing a sample of five chips. Assume 10 of the chips do not conform to customer requirements. (a) How many different samples are possible? (b) How many samples of five contain exactly one nonconforming chip? (c) How many samples of five contain at least one nonconforming chip?

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Suppose \(2 \%\) of cotton fabric rolls and \(3 \%\) of nylon fabric rolls contain flaws. Of the rolls used by a manufacturer, \(70 \%\) are cotton and \(30 \%\) are nylon. What is the probability that a randomly selected roll used by the manufacturer contains flaws?

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