/*! 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} Q. 3.30 Show that for any events Eand F,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Log out

Learning Materials

Features

Discover

Chapter 3: Q. 3.30 (page 111)

Show that for any events Eand F,

P(E∣E∪F)≥P(E∣F)

Hint: Compute P(E∣E∪F)by conditioning on whether F occurs.

Short Answer

Expert verified

The result is that P(E∣E∪F)is weighted average betweenP(E∣F)and1

Step by step solution

01

Prove

Demonstrate as all eventualities E,F

P(E∣E∪F)=P[E∣F∩(E∪F)]P(F∣E∪F)+PE∣Fc∩(E∪F)PFc∣E∪F

If it's not clear, evaluate that equations via enlarging the correct hand side by a probability density statement.

F∩(E∪F)=F

Fc∩(E∪F)=E∩Fc

⇒PE∣Fc∩(E∪F)=PE∣EFc=1

02

Weighted Average

That's also sufficient to ascertain the disparity, as

P(E∣E∪F)=P(E∣F)P(F∣E∪F)+1⋅PFc∣E∪F

P(F∣E∪F)+PFc∣E∪F=1

P(E∣F)≤1

thus localid="1649659759663" P(E∣E∪F)is that the weighted combination of such constants localid="1649659766251" P(E∣F)and localid="1649659771667" 1, so just between that two integers,

localid="1649659823011" P(E∣F)≤P(E∣E∪F)≤1

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!

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 of 2 cabinets identical in appearance has 2 drawers. Cabinet A contains a silver coin in each drawer, and cabinet B contains a silver coin in one of its drawers and a gold coin in the other. A cabinet is randomly selected, one of its drawers is opened, and a silver coin is found. What is the probability that there is a silver coin in the other drawer?


All the workers at a certain company drive to work and park in the company’s lot. The company is interested in estimating the average number of workers in a car. Which of the following methods will enable the company to estimate this quantity? Explain your answer.

1. Randomly choose n workers, find out how many were in the cars in which they were driven, and take the average of the n values.

2. Randomly choose n cars in the lot, find out how many were driven in those cars, and take the average of the n values

An urn contains 6white and9black balls. If 4balls are to be randomly selected without replacement, what is the probability that the first 2selected is white and the last 2 black?

Consider a school community of mfamilies, with niof them having ichildren, i=1,…,k,∑i=1kni=mConsider the following two methods for choosing a child:

1. Choose one of the mfamilies at random and then randomly choose a child from that family.

2. Choose one of the ∑i=1kinichildren at random.

Show that method 1is more likely than method 2to result

in the choice of a firstborn child.

Hint: In solving this problem, you will need to show that

∑i=1kini∑j=1knjj≥∑i=1kni∑j=1knj

To do so, multiply the sums and show that for all pairs i,j, the coefficient of the termninj is greater in the expression on the left than in the one on the right.

Prove the equivalence of Equations (5.11) and (5.12).
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.