/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Twenty percent of all telephones... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Understanding the problem

We are given that 20% of telephones are submitted for service, and 40% of these must be replaced. The target is to find the probability that exactly 2 out of 10 telephones need replacement.
02

Calculating Probability of a Single Telephone Being Replaced

First, we need to determine the overall probability of a single telephone being replaced. 20% of telephones are submitted, and of those, 40% are replaced. The combined probability is \(0.20 \times 0.40 = 0.08\).
03

Identifying Probability Formula

The scenario given is a binomial probability scenario where we have a fixed number of independent trials (10 telephones), two possible outcomes per trial (replaced or not), and a constant probability of a telephone needing replacement (0.08).
04

Applying the Binomial Formula

The probability of exactly \(k\) successes in \(n\) independent Bernoulli trials 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

Calculating the Binomial Probability

Using the binomial formula: \[ P(X=2) = \binom{10}{2} (0.08)^2 (0.92)^8 \]Calculate each part: - \(\binom{10}{2} = 45\)- \((0.08)^2 = 0.0064\)- \((0.92)^8 \approx 0.5132\)This results in:\[ P(X=2) = 45 \times 0.0064 \times 0.5132 = 0.1474\]
06

Conclusion

The calculated probability that exactly two telephones will be replaced out of 10 is approximately \(0.147\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Calculation
Probability calculation is an essential tool in determining the likelihood of an event happening. In the context of our exercise, we need to calculate the probability that exactly two out of ten telephones need replacement under warranty. This requires understanding both the overall probability of individual occurrences and how these contribute to the larger event.
We start by finding the probability of a single telephone being submitted and subsequently being replaced. Since 20% of telephones are submitted and only 40% of those need replacement, we calculate this by multiplying these probabilities:
  • The probability of a telephone being submitted is 0.20
  • The probability of a submitted phone being replaced is 0.40
  • The combined probability for a phone being replaced is therefore: \(0.20 \times 0.40 = 0.08\)
Understanding this calculation helps us build toward the overall probability for multiple phones, using this base probability that a single phone will be replaced.
Binomial Probability Formula
The binomial probability formula is a powerful method for calculating the probability of a specific number of successes in a series of independent and identical trials. In our case, each telephone's need for replacement is considered an independent trial.
The formula for the probability of getting exactly \(k\) replacements out of \(n\) phones is:
  • \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
Here,
  • \(n\) represents the total number of trials, which is 10 phones
  • \(k\) is the number of successes we want, which is 2 replacements
  • \(p\) is the probability of a single success, which is a phone being replaced (0.08)
  • \(1-p\) is the probability of a phone not being replaced (0.92)
To compute the exact probability, apply these values into the formula, resulting in:
  • \(\binom{10}{2} (0.08)^2 (0.92)^8\)
Understanding each component ensures accurate probability calculations in various situations.
Independent Trials
Independent trials are a fundamental concept in binomial distribution, indicating that the outcome of one trial does not impact the outcomes of another. Each trial in our phone replacement scenario is the decision whether a single phone needs replacement.
There are certain characteristics that define independent trials in binomial distributions:
  • Each trial has two possible outcomes: success (replacement) or failure (no replacement).
  • The probability of a single event (phone needing replacement) remains constant across all trials.
  • The trials themselves do not influence each other, meaning the replacement decision of one phone does not affect another.
Understanding these characteristics ensures you can accurately model and predict outcomes using binomial distribution methods—and that is exactly how we confidently determine the probability of having exactly two phones needing replacement from our batch of ten.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A mail-order computer business has six telephone lines. Let \(X\) denote the number of lines in use at a specified time. Suppose the pmf of \(X\) is as given in the accompanying table. \begin{tabular}{l|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.10\) & \(.15\) & \(.20\) & \(.25\) & \(.20\) & \(.06\) & \(.04\) \end{tabular} Calculate the probability of each of the following events. a. \\{at most three lines are in use\\} b. \\{fewer than three lines are in use\\} c. \\{at least three lines are in use\\} d. \\{between two and five lines, inclusive, are in use\\} e. \\{between two and four lines, inclusive, are not in use\\} f. \\{at least four lines are not in use\\}

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?

Suppose that \(p=P(\) male birth \()=.5\). A couple wishes to have exactly two female children in their family. They will have children until this condition is fulfilled. a. What is the probability that the family has \(x\) male children? b. What is the probability that the family has four children? c. What is the probability that the family has at most four children? d. How many male children would you expect this family to have? How many children would you expect this family to have?

Show that the cdf \(F(x)\) is a nondecreasing function; that is, \(x_{1}

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5 -lb batches. Let \(X=\) the number of batches ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.