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Chapter 8: Problem 38

QuaLm CoNTROL Refer to Exercise 37. If the manager's criterion for accepting shipment is that there be no more than 1 defective electronic game in a random sample of 20 . what is the probability that he will accept the shipment if \(10 \%\) of the electronic games is defective?

Short Answer

Expert verified
The probability that the manager will accept the shipment if 10% of the electronic games are defective is approximately 0.735 or 73.5%.

Step by step solution

01

Identify the Variables in the Problem

In this problem, we have the following variables: - n = 20 (number of trials, or electronic games in the sample) - p = 0.10 (probability of a game being defective) - k = 0 and k = 1 (the possible number of defective games in the sample that will lead to the manager accepting the shipment) We want to calculate the probability of the manager accepting the shipment, which means we need to find: \[P(0;20,0.10) + P(1;20,0.10)\]
02

Calculate the Binomial Probabilities

Let's calculate the probabilities for having 0 defective games and 1 defective game in the sample of 20. For k = 0: \[P(0;20,0.10) = \binom{20}{0}(0.10)^0(1-0.10)^{20-0}\] For k = 1: \[P(1;20,0.10) = \binom{20}{1}(0.10)^1(1-0.10)^{20-1}\]
03

Calculate the Binomial Coefficients

Now we will calculate the binomial coefficients for the probabilities above. For k = 0: \[\binom{20}{0} = \frac{20!}{0!(20-0)!} = 1\] For k = 1: \[\binom{20}{1} = \frac{20!}{1!(20-1)!} = 20\]
04

Find the Individual Probabilities

Now that we have the binomial coefficients, we can plug them back into the binomial probability equation and find the individual probabilities for k = 0 and k = 1. For k = 0: \[P(0;20,0.10) = 1(0.10)^0(0.90)^{20} = (1)(1)(0.358)\] \[P(0;20,0.10) \approx 0.358\] For k = 1: \[P(1;20,0.10) = 20(0.10)^1(0.90)^{19} = (20)(0.10)(0.398)\] \[P(1;20,0.10) \approx 0.377\]
05

Calculate the Total Probability for Accepting the Shipment

Now, we will add the individual probabilities to get the total probability that the manager will accept the shipment. \[P = P(0;20,0.10) + P(1;20,0.10) \] \[P \approx 0.358 + 0.377 = 0.735\] The probability that the manager will accept the shipment if 10% of the electronic games are defective is approximately 0.735 or 73.5%.

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Key Concepts

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

Probability Calculation
When dealing with probability, particularly in situations involving multiple independent events such as defective items in a shipment, we use specific methods to determine outcomes.
One of the foundational elements in such a calculation is understanding what probability means: it is essentially the measure of the likelihood that a particular event will occur.
In the context of our problem scenario, we calculate the probability of having no more than a certain number of defective items in a sample.
  • The probability of an individual defective item is given as 10% or 0.10.
  • We aim to compute the likelihood that a random sample of 20 items contains at most 1 defective item.
This type of calculation requires the use of binomial probability, which helps when each trial is independent and there are only two possible outcomes: defective or not defective.
Defective Items Analysis
Analyzing defective items involves determining how many defective items are permissible for the manager to accept a shipment.
This analysis is crucial for quality control in manufacturing and product shipment scenarios.
In this problem, the manager accepts the shipment if there are zero or only one defective item in a random sample of 20 products.
  • "Defective items analysis" includes understanding the tolerance limit for defects.
  • It guides decision-making to ensure quality remains within standards.
Tracking the number of defective items helps maintain product quality and customer satisfaction.
In our case, we calculated probabilities for having exactly 0 or 1 defective items, which involves understanding how likely both scenarios are when taking into account the known defect rate of products.
Binomial Coefficients
To accurately calculate the probabilities of different numbers of defective items, we use binomial coefficients.
The binomial coefficient, denoted as \(\binom{n}{k}\), is a factor that plays an important role in determining the weight of a specific outcome in a binomial probability distribution.
  • The formula for a binomial coefficient is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of trials, and \(k\) is the number of successful outcomes you are calculating.
  • In our scenario, we calculate the number of ways to choose 0 or 1 defective item from 20, which are \(\binom{20}{0} = 1\) and \(\binom{20}{1} = 20\).
These coefficients help to weigh the probabilities of specific configurations of defective and non-defective items.
By computing these coefficients, we are then able to accurately plug them into the binomial probability formula, enhancing our understanding of potential outcomes.

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Most popular questions from this chapter

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