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Chapter 3: Problem 55

Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

Short Answer

Expert verified
The probability is approximately 0.147.

Step by step solution

01

Understand the Problem

We know that 20% of the telephones are submitted for service under warranty, and of those, 40% must be replaced. Thus, the probability that a telephone needs to be replaced is the product of being serviced and needing replacement. Let's calculate this probability.
02

Calculate Probability of Replacement

The probability that a telephone is replaced is the probability of being serviced (0.20) times the probability of being replaced once serviced (0.40). So, the probability, \( p \), that a telephone is replaced is: \[ p = 0.20 \times 0.40 = 0.08 \]
03

Identify the Binomial Distribution

This is a binomial problem because we are finding the probability of a certain number of successes (telephones being replaced) in a fixed number of trials (10 telephones). Here, the probability of success on each trial (a phone needing replacement) is 0.08, and we are considering 10 trials.
04

Setup the Binomial Probability Formula

The probability of exactly \( k \) successes in \( n \) trials for a binomial distribution is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( n = 10 \), \( k = 2 \), and \( p = 0.08 \).
05

Calculate the Binomial Coefficient

Calculate the binomial coefficient \( \binom{10}{2} \):\[ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
06

Evaluate the Probability

Calculate \( P(X = 2) \) using the formula:\[ P(X = 2) = 45 \times (0.08)^2 \times (0.92)^8 \]
07

Insert Numerical Values and Calculate the Result

Substitute and calculate:\[ P(X = 2) = 45 \times 0.0064 \times 0.5132 = 0.14735 \] After performing the calculation, the probability that exactly two phones will need replacement is approximately 0.147.

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Key Concepts

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

Probability Calculation
In probability, determining the likelihood of an event involves understanding the mechanics of random occurrences. In our exercise, we're interested in the probability that exactly two out of ten phones require replacement under warranty.
This requires first calculating the individual replacement probability of each phone.
  • The probability of a phone being serviced is 0.20.
  • The probability of a phone needing replacement once serviced is 0.40.
  • Thus, the overall probability of a phone being replaced is the product of these two individual probabilities, calculated as:
    \( p = 0.20 \times 0.40 = 0.08 \).
This probability (0.08) is crucial to our main exercise and sets the stage for the binomial distribution usage. Such calculations underpin much of what follows in analyzing how likely it is that a specific number of replacements occur.
Replacement Probability
Understanding replacement probability deals with scenarios familiar in quality control and warranty analysis. Here, replacement probability refers to the chance that an individual item, once checked, fails to meet operational criteria and thus, must be substituted.
  • First, identify the chance that an item will be checked, represented by servicing probability (0.20 in this case).
  • Then, isolate the further step where such a checked item will need replacing (40%).
  • By multiplying these probabilities, you derive the replacement probability. For this phone exercise, it is fairly low at 0.08.
Through this, replacement probability helps in planning resources for replacements and estimating customer service needs. It is a critical factor in logistic and financial planning in any service or manufacturing industry dealing with warranties.
Binomial Coefficient
The binomial coefficient is a fundamental concept when working with binomial probabilities. It represents the number of ways to choose a certain number of successes from a pool of trials. In our exercise, we're working with 10 phones and want 2 replacements, so we need to calculate the binomial coefficient for these numbers.
  • Use the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). For 10 phones and 2 replacements, it becomes:
  • \( \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \)
Once calculated, this binomial coefficient is combined with the replacement probability in the binomial probability formula. This allows you to find the probability of exactly two replacements occurring. Thus, understanding and using the binomial coefficient ensures you can predict exact outcomes within a set sample using probabilistic methods.

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Most popular questions from this chapter

A company that produces fine crystal knows from experience that \(10 \%\) of its goblets have cosmetic flaws and must be classified as "seconds." a. Among six randomly selected goblets, how likely is it that only one is a second? b. Among six randomly selected goblets, what is the probability that at least two are seconds? c. If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?

The article "Reliability-Based Service-Life Assessment of Aging Concrete Structures" (J. Structural Engr., 1993: 1600-1621) suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is . 5 year. a. How many loads can be expected to occur during a 2-year period? b. What is the probability that more than five loads occur during a 2 -year period? c. How long must a time period be so that the probability of no loads occurring during that period is at most .1?

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Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\mu=20\) (suggested in the article "'Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Each time a component is tested, the trial is a success ( \(S\) ) or failure \((F)\). Suppose the component is tested repeatedly until a success occurs on three consecutive trials. Let \(Y\) denote the number of trials necessary to achieve this. List all outcomes corresponding to the five smallest possible values of \(Y\), and state which \(Y\) value is associated with each one.
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