/*! 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 17 Divide a line segment into two p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 17

Divide a line segment into two parts by selecting a point at random. Find the probability that the larger segment is at least three times the shorter. Assume a uniform distribution.

Short Answer

Expert verified
The probability that the larger segment is at least three times as long as the smaller segment when a line segment is divided at a random point is \(\frac{1}{2}\).

Step by step solution

01

Define the Problem

Let the whole length of the line be '1'. Hence, if a point is selected at random, the length of the segment from the chosen point to one end will range from 0 to 1. Let that length be \(x\). Therefore, the length of the remaining segment (1-\(x\)) is three times \(x\). This can be written as \(x \leq \frac{1}{4}\) and \(x \geq \frac{3}{4}\)
02

Calculate Probability

The total probability space is '1' because a point can range from '0' to '1'. To find the probability that the larger segment is at least three times the length of the smaller one, add the probabilities of two disjoint ranges, \([0, \frac{1}{4}]\) and \([\frac{3}{4}, 1]\). Each range has a length of \(\frac{1}{4}\). The sum of these gives a total probability of \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\). Thus, the probability that the larger segment is three times the smaller one is \(\frac{1}{2}\).

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 91影视!

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

Let a random variable \(X\) of the continuous type have a pdf \(f(x)\) whose graph is symmetric with respect to \(x=c\). If the mean value of \(X\) exists, show that \(E(X)=c\) Hint: Show that \(E(X-c)\) equals zero by writing \(E(X-c)\) as the sum of two integrals: one from \(-\infty\) to \(c\) and the other from \(c\) to \(\infty\). In the first, let \(y=c-x\); and, in the second, \(z=x-c .\) Finally, use the symmetry condition \(f(c-y)=f(c+y)\) in the first.

Let \(X\) have the exponential pdf, \(f(x)=\beta^{-1} \exp \\{-x / \beta\\}, 0

From a bowl containing 5 red, 3 white, and 7 blue chips, select 4 at random and without replacement. Compute the conditional probability of 1 red, 0 white, and 3 blue chips, given that there are at least 3 blue chips in this sample of 4 chips.

Find the pdf \(f(x)\), the 25 th percentile, and the 60 th percentile for each of the following cdfs: Sketch the graphs of \(f(x)\) and \(F(x)\). (a) \(F(x)=\left(1+e^{-x}\right)^{-1},-\infty

A bowl contains ten chips numbered \(1,2, \ldots, 10\), respectively. Five chips are drawn at random, one at a time, and without replacement. What is the probability that two even-numbered chips are drawn and they occur on even- numbered draws?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
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.