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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 0.1472.

Step by step solution

01

Understand the Problem

We want to find the probability that exactly two out of ten purchased telephones will need to be replaced under warranty. We know that 20% of phones need service and 40% of those need replacement.
02

Calculate Overall Replacement Probability for One Phone

Calculate the probability that a phone needs to be replaced. If 20% are submitted for service and 40% of these need replacement, the probability a phone will be replaced is \(0.2 \times 0.4 = 0.08\).
03

Define the Binomial Experiment

Purchasing 10 phones is a binomial experiment where each trial has a success probability of 0.08 (a phone being replaced). We want the probability of exactly 2 successes (replacements) out of 10 trials.
04

Apply Binomial Probability Formula

The binomial probability formula 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 Combinatorial Factor

Calculate \(\binom{10}{2}\): \[\binom{10}{2} = \frac{10!}{2!(10-2)!} = 45\]
06

Calculate Probability for 2 Replacements

Substitute values into the binomial probability formula:\[P(X = 2) = 45 \times (0.08)^2 \times (0.92)^8\]Calculate each component step-by-step.
07

Evaluate Probability Components

Calculate \((0.08)^2 = 0.0064\) and \((0.92)^8 = 0.5132\). Then multiply these with the combinatorial factor:\[ P(X = 2) = 45 \times 0.0064 \times 0.5132 = 0.1472\]
08

Conclude the Solution

Therefore, the probability that exactly two out of ten purchased telephones will be replaced is approximately 0.1472.

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Key Concepts

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

Binomial Distribution
The Binomial Distribution is a key concept in probability theory used to model situations where there are a fixed number of independent trials, each with two possible outcomes -- often referred to as "success" and "failure." In our exercise, buying and evaluating the warranty status of each phone represents one trial, with the outcomes being either the phone is replaced (success) or it's not (failure).

Key aspects of a binomial distribution include:
  • Finite number of trials (e.g., 10 phones)
  • Each trial is independent of others
  • Two possible outcomes per trial (replacement or no replacement)
  • Constant probability of success in each trial (replacing a phone)

The probability of achieving exactly a certain number of successes (e.g., exactly two phones needing replacement) can be calculated using the binomial probability formula. This involves determining the number of ways the desired number of successes can occur (combinations) and adjusting for the respective probabilities of success and failure.
Combinatorial Factor
The concept of the Combinatorial Factor is essential when calculating probabilities involving multiple events or trials. In the context of the binomial distribution, it allows us to determine how many different ways a certain number of "successes" can be achieved within a series of trials.

For example, in our exercise, we need to find how many different ways two phones out of ten can be replaced. This is symbolized as \( \binom{10}{2} \), representing the number of combinations for choosing 2 successes out of 10 trials.

The formula to find the combinatorial factor is:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where:
  • \( n \) is the total number of trials
  • \( k \) is the number of successful trials
  • \( ! \) denotes the factorial function, which multiplies a sequence of descending natural numbers

In our example, \( \binom{10}{2} \) is calculated as \( \frac{10!}{2!(10-2)!} = 45 \), meaning there are 45 different combinations where two of the ten phones need replacement. This factor is crucial for the binomial probability formula.
Warranty Service Probability
Warranty Service Probability is the likelihood that a product, like a telephone, is repaired or replaced under its warranty terms. This probability is crucial for manufacturers and consumers alike in assessing product reliability and expected service rates.

In the given exercise, each phone has a probability of being replaced which is derived from known facts:
  • 20% of the phones require service. This is the percentage of phones that need any sort of warranty intervention.
  • Of these, 40% will end up being replaced. This subset focuses on products that cannot be repaired and must be replaced.
Thus the total probability of a phone needing replacement under warranty is the product of these probabilities: \(0.2 \times 0.4 = 0.08\).

Applying this probability uniformly across multiple devices, as in the exercise where 10 phones are purchased, forms the backbone of the binomial distribution model and helps in calculating the probability of exactly two replacements. Understanding this combination of probabilities gives insight into managing expectations of product performance and necessary reserves for warranty services.

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Most popular questions from this chapter

Consider a deck consisting of seven cards, marked \(1,2, \ldots, 7\). Three of these cards are selected at random. Define an rv \(W\) by \(W=\) the sum of the resulting numbers, and compute the pmf of \(W\). Then compute \(\mu\) and \(\sigma^{2}\). [Hint: Consider outcomes as unordered, so that \((1,3,7)\) and \((3,1,7)\) are not different outcomes. Then there are 35 outcomes, and they can be listed. (This type of rv actually arises in connection with Wilcoxon's rank-sum test, in which there is an \(x\) sample and a \(y\) sample and \(W\) is the sum of the ranks of the \(x^{+}\)s in the combined sample.)]

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