/*! 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 57 Suppose that \(90 \%\) of all ba... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(90 \%\) of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?

Short Answer

Expert verified
The probability that at least nine will work is 0.4576, assuming independent voltages.

Step by step solution

01

Understanding the Problem

We need to find the probability that at least nine out of ten flashlights will work, given that each flashlight requires two batteries to work, and each battery has a 90% chance of having an acceptable voltage. For a flashlight to work, both batteries must be acceptable.
02

Determining Probability of a Flashlight Working

For a flashlight to work, both batteries must have acceptable voltages. Since each battery has a 0.9 probability of being acceptable, the probability that a flashlight works is \[(0.9)^2 = 0.81\].This is because the flashlight will work if the first battery works (0.9 probability) AND the second battery works (0.9 probability), resulting in the product 0.9 * 0.9.
03

Identify Probability Distribution

The problem of determining the probability of a given number of flashlights working among ten draws on the binomial distribution, where the probability of a flashlight working is 0.81, and we are selecting 10 flashlights.
04

Binomial Probability Formula

The probability of exactly \(k\) successes (working flashlights) out of \(n\) trials is given by the binomial probability formula: \[P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k}\] where \(n = 10\), \(k\) is the number of working flashlights, and \(p = 0.81\).
05

Calculating Probability for 9 Working Flashlights

Using the formula, calculate the probability for exactly 9 working flashlights (\(k = 9\)):\[P(X = 9) = \binom{10}{9} \times (0.81)^9 \times (0.19)^1\]\[= 10 \times 0.177978 \times 0.19 \approx 0.336\]
06

Calculating Probability for 10 Working Flashlights

Similarly, calculate the probability for exactly 10 working flashlights (\(k = 10\)):\[P(X = 10) = \binom{10}{10} \times (0.81)^{10} \times (0.19)^0\]\[= 1 \times 0.12157665459 \times 1 = 0.1216\]
07

Summing Probabilities for At Least 9

The probability that at least 9 flashlights work is the sum of the probabilities of exactly 9 working and exactly 10 working:\[P(X \geq 9) = P(X = 9) + P(X = 10)\]\[= 0.336 + 0.1216 = 0.4576\]
08

Stating Assumptions

The assumption here is that the condition (each battery has a 90% probability of being acceptable) is independent for each battery, and the sample of flashlights is randomly selected.

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Key Concepts

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

Probability Calculation
Understanding how to calculate probability is essential when determining the likelihood of a particular event happening. Probability is the measure of how likely something is to occur, ranging from 0 (impossible) to 1 (certain). To calculate probability, especially in scenarios like our battery example, we can follow a simple approach:
  • Identify the total number of possible outcomes.
  • Determine the number of favorable outcomes.
  • Divide the number of favorable outcomes by the total number of possible outcomes.
This forms the basic formula for probability: \[ P = \frac{\text{Favorable Outcomes}}{\text{Total Possible Outcomes}} \]
In more complex problems, like our flashlight example, understanding the sequence and combination of independent outcomes (like batteries having a 90% chance) will guide you to use the appropriate method—such as binomial distribution, which we'll talk about next.
Binomial Probability Formula
When you're dealing with situations where a fixed number of independent experiments (trials) are conducted, such as checking if a flashlight works, the binomial probability formula becomes useful. This formula helps to calculate the probability of having a certain number of successful outcomes in a series of independent events. Let's break it down:1. **Parameters**: - **n**: the total number of trials or experiments (in our case, 10 flashlights). - **k**: the number of successful trials (either 9 or 10 flashlights working). - **p**: the probability of success on a single trial (0.81 for a flashlight working).The formula is expressed as:\[ P(X = k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \]Where:- \( \binom{n}{k} \) symbolizes combinations and refers to the number of ways to choose \( k \) successes from \( n \) trials.- \( p^k \): probability of having exactly \( k \) successes.- \( (1-p)^{n-k} \): probability of having \( n-k \) failures.
For our flashlight example, applying this formula enables us to calculate the chances of exactly 9 or 10 flashlights working, helping us sum those probabilities to determine the overall likelihood of at least 9 working flashlights.
Independent Events
In probability theory, independent events are crucial because the outcome of one event does not affect the outcome of another. This assumption simplifies calculations significantly and is fundamental to applying methods like the binomial probability formula.When we say events are independent:- The occurrence of one does not change the likelihood of the other occurring.- Mathematically, two events A and B are independent if: \[ P(A \cap B) = P(A) \times P(B) \]
In the flashlight exercise, the acceptable voltage of each battery is considered an independent event. This means that whether a battery is good or not does not impact the status of another. The independence allows us to multiply probabilities for serial events—such as calculating that a flashlight works only if both batteries do. Hence, the probability of a flashlight working is derived by multiplying the probability of each battery having acceptable voltage (0.9 for each), leading to the result \( (0.9)^2 \ = 0.81 \), establishing a foundation for further probability calculations.

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Most popular questions from this chapter

The pmf of the amount of memory \(X(G B)\) in a purchased flash drive was given in Example \(3.13\) as \begin{tabular}{l|lllll} \(x\) & 1 & 2 & 4 & 8 & 16 \\ \hline\(p(x)\) & \(.05\) & \(.10\) & \(.35\) & \(.40\) & \(.10\) \end{tabular} Compute the following: a. \(E(X)\) b. \(V(X)\) directly from the definition c. The standard deviation of \(X\) d. \(V(X)\) using the shortcut formula

A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2 . a. What is the probability that the batch will be accepted when the actual proportion of defectives is .01? .05?.10? 20 ?.25? b. Let \(p\) denote the actual proportion of defectives in the batch. A graph of \(P\) (batch is accepted) as a function of \(p\), with \(p\) on the horizontal axis and \(P\) (batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for \(0 \leq p \leq 1\). c. Repeat parts (a) and (b) with " 1 " replacing " 2 " in the acceptance sampling plan. d. Repeat parts (a) and (b) with " 15 " replacing " \(10 "\) in the acceptance sampling plan. e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

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?

An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let \(p=\) the true proportion of such houses having detectors, and suppose that a random sample of 25 homes is inspected. If the sample strongly indicates that fewer than \(80 \%\) of all houses have a detector, the fire department will campaign for a mandatory inspection program. Because of the costliness of the program, the department prefers not to call for such inspections unless sample evidence strongly argues for their necessity. Let \(X\) denote the number of homes with detectors among the 25 sampled. Consider rejecting the claim that \(p \geq .8\) if \(x \leq 15\) a. What is the probability that the claim is rejected when the actual value of \(p\) is 8 ? b. What is the probability of not rejecting the claim when \(p=.7 ?\) When \(p=.6 ?\) c. How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14 ?

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