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Chapter 2: Problem 135

Eight cavities in an injection-molding tool produce plastic connectors that fall into a common stream. A sample is chosen every several minutes. Assume that the samples are independent. (a) What is the probability that five successive samples were all produced in cavity one of the mold? (b) What is the probability that five successive samples were all produced in the same cavity of the mold? (c) What is the probability that four out of five successive samples were produced in cavity one of the mold?

Short Answer

Expert verified
(a) \( \frac{1}{32768} \); (b) \( \frac{1}{4096} \); (c) \( \frac{35}{32768} \).

Step by step solution

01

Identify Basic Probability of One Sample from Cavity One

There are 8 cavities, and each sample is equally likely to come from any of these cavities. Therefore, the probability that a single sample comes from cavity one is given by \( \frac{1}{8} \).
02

Calculate Probability for Part (a)

For five successive samples to all come from cavity one, we need to multiply the probability of one sample by itself five times:\[\left( \frac{1}{8} \right)^5 = \frac{1}{32768}.\]Thus, the probability is \( \frac{1}{32768} \).
03

Identify Probability for Any Single Cavity (Part b)

Each sample can come from any one of the 8 cavities, and we want them to all come from the same cavity over five samples. The probability for a single cavity is \( \left( \frac{1}{8} \right)^5 = \frac{1}{32768} \).
04

Use Symmetry in the Problem for Part (b)

Since there are 8 identical cavities, the event that five successive samples are from the same cavity could occur for any of the cavities. So, multiply the probability of choosing one specific cavity by 8:\[8 \times \frac{1}{32768} = \frac{1}{4096}.\]
05

Compute Probability for Four Samples from Cavity One (Part c)

Using the binomial probability formula, we want 4 out of 5 samples to be from cavity one:\[\binom{5}{4} \left( \frac{1}{8} \right)^4 \left( \frac{7}{8} \right)^1.\]Calculate each component:- \( \binom{5}{4} = 5 \)- \( \left( \frac{1}{8} \right)^4 = \frac{1}{4096} \)- \( \left( \frac{7}{8} \right)^1 = \frac{7}{8} \).Thus, the probability is:\[5 \times \frac{1}{4096} \times \frac{7}{8} = \frac{35}{32768}.\]
06

Simplify and Verify Final Results

Double-check calculations and ensure probabilities are simplified:1. Probability for (a) is \( \frac{1}{32768} \).2. Probability for (b) is \( \frac{1}{4096} \), after multiplying by 8.3. Probability for (c) is \( \frac{35}{32768} \), using the binomial formula.

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Key Concepts

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

Injection Molding
Injection molding is a manufacturing process used to create parts by injecting material into a mold. In this context, we are discussing a mold with eight cavities, each contributing to producing plastic connectors.
The connectors from each cavity flow into the same stream, making it difficult to discern which cavity produced which part, unless an analysis is performed.
  • The mold's cavities function like independent machines, each having an equal chance of producing the next connector in the stream.
  • This setup introduces randomness in determining the origin of each sample connector.
  • The cavities could potentially affect the probability distribution of where samples come from, crucial for understanding the given problem.
This process simulates a setting often analyzed in probability theory to consider outcomes like those calculated in the exercise.
Binomial Distribution
The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states across a specific number of experiments. In the context of the injection-molding problem, this distribution is applicable because each sample either comes from cavity one or does not.
  • Each connector sample is a trial, with a fixed number of eight cavities, implying a total of two potential states: from cavity one or from another cavity.
  • The likelihood of a sample emerging from cavity one (success) is constant, emphasizing this as a Bernoulli trial setup. These trials are repeated multiple times depending on how many connectors are examined.
  • The properties of binomial distribution allow us to use formulas to calculate the probabilities of events like getting four out of five samples from cavity one, where multiple samples may meet the success criterion, making it a perfect fit for the problem.
This helps us conceptualize how likely certain outcomes are during the sampled trials.
Independent Events
In probability theory, independent events are those whose outcomes do not influence each other. For the problem at hand, it is assumed that the samples taken at every interval are independent of previous samples.
Because these samples are independent:
  • The occurrence of a connector coming from a specific cavity does not affect the next one.
  • This means knowing the source of one sample gives us no additional information about subsequent samples.
  • Independence simplifies the calculations of probabilities across multiple samples, as one can directly multiply the probability of individual events.
This characteristic underpins the calculations we use in solving the probability queries associated with the injection molding problem.
Sample Analysis
Sample analysis in probability involves determining the likelihood of various outcomes from samples taken. In this situation, a sample analysis is used to assess the probability that successive samples originate from specific cavities.
Important considerations include:
  • Understanding that each sampled connector represents a random draw from the population of total connectors produced.
  • Through analyzing samples, one can infer the probabilities of particular events, like multiple successive connectors from the same cavity, rather than just a single case.
  • Such analysis requires the use of probability theory techniques, like those discussed in the step-by-step solution, including symmetry and binomial probability calculations.
Sample analysis helps confirm predictions about production habits within the injection molding process.

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Most popular questions from this chapter

The sample space of a random experiment is \(\\{a, b, c\), \(d, e\\}\) with probabilities \(0.1,0.1,0.2,0.4,\) and \(0.2,\) respectively. Let \(A\) denote the event \(\\{a, b, c\\},\) and let \(B\) denote the event \(\\{c, d, e\\} .\) Determine the following: (a) \(P(A)\) (b) \(P(B)\) (c) \(P\left(A^{\prime}\right)\) (d) \(P(A \cup B)\) (e) \(P(A \cap B)\)

In circuit testing of printed circuit boards, each board either fails or does not fail the test. A board that fails the test is then checked further to determine which one of five defect types is the primary failure mode. Represent the sample space for this experiment.

In a magnetic storage device, three attempts are made to read data before an error recovery procedure that repositions the magnetic head is used. The error recovery procedure attempts three repositionings before an "abort" message is sent to the operator. Let \(s\) denote the success of a read operation \(f\) denote the failure of a read operation \(F\) denote the failure of an error recovery procedure \(S\) denote the success of an error recovery procedure \(A\) denote an abort message sent to the operator. Describe the sample space of this experiment with a tree diagram.

In the design of an electromechanical product, 12 components are to be stacked into a cylindrical casing in a manner that minimizes the impact of shocks. One end of the casing is designated as the bottom and the other end is the top. (a) If all components are different, how many different designs are possible? (b) If seven components are identical to one another, but the others are different, how many different designs are possible? (c) If three components are of one type and identical to one another, and four components are of another type and identical to one another, but the others are different, how many different designs are possible?

A sample of three calculators is selected from a manufacturing line, and each calculator is classified as either defective or acceptable. Let \(A, B,\) and \(C\) denote the events that the first, second, and third calculators, respectively, are defective. (a) Describe the sample space for this experiment with a tree diagram. Use the tree diagram to describe each of the following events: (b) \(A\) (c) \(B\) (d) \(A \cap B\) (e) \(B \cup C\)
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