/*! 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 14 Suppose that you have ten lightb... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter \(\lambda\). a. What is the probability that all ten bulbs fail before time \(r\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(t\) ? c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \(\lambda\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \(\theta\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time \(t\) ?

Short Answer

Expert verified
a. Probability is \((1-e^{-\lambda r})^{10}\). b. Probability is \(\binom{10}{k}(1-e^{-\lambda t})^k(e^{-\lambda t})^{10-k}\). c. Probability involves the non-standard bulb in any 5 failures with adjusted rates.

Step by step solution

01

Calculate Probability for Single Bulb

The probability that a single light bulb with an exponential distribution fails before time \( r \) is given by the cumulative distribution function (CDF) of the exponential distribution: \( P(X < r) = 1 - e^{-\lambda r} \).
02

Probability All Bulbs Fail Before Time r

For all ten bulbs to fail before time \( r \), we need each bulb to fail independently before \( r \). Therefore, the probability that all ten bulbs fail is: \(P(X_1 < r, X_2 < r, \ldots, X_{10} < r) = \left(1 - e^{-\lambda r}\right)^{10}\).
03

Use Binomial Distribution for Part b

Since each bulb fails independently, and failure before time \( t \) for a given bulb is a Bernoulli trial, the number of bulbs failing before time \( t \) follows a Binomial distribution. The probability of exactly \( k \) bulbs failing is given by: \( P(\text{exactly } k \text{ failures}) = \binom{10}{k} \left(1 - e^{-\lambda t}\right)^k \left(e^{-\lambda t}\right)^{10-k} \).
04

Modify Distribution for Part c

In Part (c), one bulb has a different parameter (\(\theta\)). Let the nine bulbs use a success probability of \((1-e^{-\lambda t})\), and the special bulb with \((1-e^{-\theta t})\). The distribution is no longer straightforwardly binomial but can be handled by considering combinations. Let's find the probability exactly five bulbs, including the special bulb, fail before time \( t \): \( \sum_{k=0}^{5} \binom{9}{k}\left(1 - e^{-\lambda t}\right)^{k} \left(e^{-\lambda t}\right)^{9-k} \cdot \left(1 - e^{-\theta t}\right)\). This sums over cases where exactly one fixed different bulb among failures.
05

Calculate Composite Probability for Part c

The total probability that exactly five bulbs fail, including the one with \(\theta\) as a failure, involves adding probabilities of valid combinations of 5 failures (where one of them is the non-standard bulb). Essentially, compute: \(P(5\text{ failures}) = \binom{9}{4} \times (1-e^{-\lambda t})^4 e^{-\lambda t}^{5} \times (1-e^{-\theta t})\).

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.

Exponential Distribution
The exponential distribution is a continuous probability distribution often used to model the time or space between events in a Poisson process. It is characterized by a parameter \( \lambda \), known as the rate parameter. This parameter helps determine how quickly an event occurs. The exponential distribution is particularly useful for modeling scenarios where things "wear out" or "fail," like the lifespan of a lightbulb.
Understanding the basic properties of this distribution is key:
  • The mean of an exponentially distributed random variable \( X \) is \( E(X) = \frac{1}{\lambda} \).
  • The variance is \( \text{Var}(X) = \frac{1}{\lambda^2} \).
  • The probability that the event happens before time \( t \) is given by the cumulative distribution function (CDF): \( P(X < t) = 1 - e^{-\lambda t} \).
In the context of lightbulbs, an exponential distribution helps us determine the probability of failure within a certain time, considering each lightbulb independently of others.
Binomial Distribution
The binomial distribution is a discreet probability distribution of the number of successes in a sequence of \( n \) independent experiments or trials, each asking a yes-no question. Such questions can either result in "success" or "failure." The distribution is defined by two parameters: \( n \), the number of trials, and \( p \), the probability of success on each trial.
The formula for finding the probability of exactly \( k \) successes (failures in our exercise with lightbulbs) is:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
This is where \( \binom{n}{k} \) is a combination that represents the number of ways to choose \( k \) successes from \( n \) trials.
In our exercise, when calculating the number of bulbs that fail before time \( t \), we use the cumulative probability that a single bulb fails before time \( t \) as our "success" probability \( p \), while \( n \) stands for the ten bulbs collectively being tested.
Independent Events
In probability theory, events are defined as independent if the occurrence of one event does not affect the likelihood of the occurrence of another. This concept simplifies the calculation of probabilities in complex scenarios since you can treat each event separately.
In mathematical terms, two events \( A \) and \( B \) are independent if:
  • \( P(A \cap B) = P(A) \cdot P(B) \)
For our context, when calculating the probability of independent lightbulbs failing, knowing the result of one bulb's lifetime does not impact or inform us about others. This gives us the freedom to consider each bulb's failure individually when computing probabilities, such as all bulbs failing before a given time \( r \).
Recognizing the independence of events allows us to combine individual probabilities to find the total probability in a straightforward manner.

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

Show that if \(Y=a X+b(a \neq 0)\), then \(\operatorname{Corr}(X, Y)=+1\) or \(-1\). Under what conditions will \(\rho=+1\) ?

It is known that \(80 \%\) of all brand A zip drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that \(n=10\) drives are randomly selected. Let \(X=\) the number of successes in the sample. The statistic \(X / n\) is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value of \(X / n\) is \(.3\), corresponding to \(X=3\). What is the probability of this value (what kind of random variable is \(X\) )?]

Suppose the amount of liquid dispensed by a certain machine is uniformly distributed with lower limit \(A=8 \mathrm{oz}\) and upper limit \(B=10 \mathrm{oz}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes \(n=5,10,20\), and 30 .

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in 1000 s of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1\) )? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and 2 ?

Annie and Alvie have agreed to meet for lunch between noon ( \((0: 00 \mathrm{p} . \mathrm{M} .)\) and 1:00 P.M. Denote Annie's arrival time by \(X\), Alvie's by \(Y\), and suppose \(X\) and \(Y\) are independent with pdf's $$ \begin{aligned} &f_{X}(x)=\left\\{\begin{array}{cl} 3 x^{2} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \\ &f_{Y}(y)=\left\\{\begin{array}{rl} 2 y & 0 \leq y \leq 1 \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: \(h(X, Y)=\) \(|X-Y| .]\)

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.