/*! 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 Let \(X\) have the pdf \(f(x)=2 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 14

Let \(X\) have the pdf \(f(x)=2 x, 0

Short Answer

Expert verified
The probability that \(X\) is at least \(\frac{3}{4}\) given that \(X\) is at least \(\frac{1}{2}\) is calculated by the ratio of the probability that \(X\) is at least \(\frac{3}{4}\) and the probability that \(X\) is at least \(\frac{1}{2}\). It is the result of the calculation in Step 4.

Step by step solution

01

Identify the Probability Density Function

The given probability density function is \(f(x) = 2x\) within the interval \(0 < x < 1\), and zero elsewhere. This function is a pdf because it satisfies the two conditions for a pdf: firstly, \(f(x) \geq 0 \) for all \(x\), and secondly, the area under the curve over all possible values of \(x\) is 1. This can be confirmed by integrating \(f(x) = 2x\) from 0 to 1.
02

Find the Denominator of the Conditional Probability

The denominator of the conditional probability is the probability that \(X\) is at least \(\frac{1}{2}\). This can be found by integrating the pdf from \(\frac{1}{2}\) to 1:\[\int_{\frac{1}{2}}^{1} 2x \, dx\]
03

Find the Numerator of the Conditional Probability

The numerator of the conditional probability is the probability that \(X\) is at least \(\frac{3}{4}\). This can be found by integrating the pdf from \(\frac{3}{4}\) to 1:\[\int_{\frac{3}{4}}^{1} 2x \, dx\]
04

Compute the Conditional Probability

Finally, we can calculate the conditional probability as the quotient of the numerator (probability that \(X\) is at least \(\frac{3}{4}\)) and the denominator (probability that \(X\) is at least \(\frac{1}{2}\)).

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

Each bag in a large box contains 25 tulip bulbs. It is known that \(60 \%\) of the bags contain bulbs for 5 red and 20 yellow tulips, while the remaining \(40 \%\) of the bags contain bulbs for 15 red and 10 yellow tulips. A bag is selected at random and a bulb taken at random from this bag is planted. (a) What is the probability that it will be a yellow tulip? (b) Given that it is yellow, what is the conditional probability it comes from a bag that contained 5 red and 20 yellow bulbs?

Let \(X\) be a random variable with space \(\mathcal{D}\). For \(D \subset \mathcal{D}\), recall that the probability induced by \(X\) is \(P_{X}(D)=P[\\{c: X(c) \in D\\}] .\) Show that \(P_{X}(D)\) is a probability by showing the following: (a) \(P_{X}(\mathcal{D})=1\). (b) \(P_{X}(D) \geq 0\). (c) For a sequence of sets \(\left\\{D_{n}\right\\}\) in \(\mathcal{D}\), show that $$ \left\\{c: X(c) \in \cup_{n} D_{n}\right\\}=\cup_{n}\left\\{c: X(c) \in D_{n}\right\\} $$ (d) Use part (c) to show that if \(\left\\{D_{n}\right\\}\) is sequence of mutually exclusive events, then $$ P_{X}\left(\cup_{n=1}^{\infty} D_{n}\right)=\sum_{n=1}^{\infty} P\left(D_{n}\right) $$

Let the probability set function of the random variable \(X\) be $$ P_{X}(C)=\int_{C} e^{-x} d x, \quad \text { where } \mathcal{C}=\\{x: 0

Let \(X\) be a random variable such that \(E\left[(X-b)^{2}\right]\) exists for all real \(b\). Show that \(E\left[(X-b)^{2}\right]\) is a minimum when \(b=E(X)\).

If the variance of the random variable \(X\) exists, show that $$ E\left(X^{2}\right) \geq[E(X)]^{2} $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

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.