/*! 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} Q6E Suppose that X has the geometric... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Q6E (page 302)

Suppose that X has the geometric distribution with parameter p. Determine the probability that the value ofX will be one of the even integers 0, 2, 4, . . . .

Short Answer

Expert verified

\[\sum\limits_{i = 0}^\infty {P\left[ {X = 2i} \right]} = \frac{1}{{2 - p}}\]

Step by step solution

01

Given information

Here, X is a random variable that follows a geometric distribution\(X \sim Geom\left( p \right)\).

02

Find the pdf of the distribution

\[P\left( {X = k} \right) = {\left( {1 - p} \right)^k} \times p,k = 0,1,2, \ldots \]

03

To find the probability that X takes even values.

\[\begin{array}{c}\sum\limits_{i = 0}^\infty {P\left[ {X = 2i} \right]} \\ = \sum\limits_{i = 0}^\infty {p{{\left( {1 - p} \right)}^{2i}}} \\ = p\sum\limits_{i = 0}^\infty {{{\left( {1 - p} \right)}^{2i}}} \\ = \frac{p}{{1 - {{\left( {1 - p} \right)}^2}}}\\ = \frac{1}{{2 - p}}\end{array}\]

Hence the answer is\[\frac{1}{{2 - p}}\]

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

Suppose that an electronic system contains n components that function independently of each other, and suppose that these components are connected in series, as defined in Exercise 5 of Sec. 3.7. Suppose also that each component will function properly for a certain number of periods and then will fail. Finally, suppose that for i =1,...,n, the number of periods for which component i will function properly is a discrete random variable having a geometric distribution with parameter \({p_i}\). Determine the distribution of the number of periods for which the system will function properly.

Find the value of \(\left( {\begin{array}{*{20}{c}}{\frac{3}{2}}\\4\end{array}} \right)\).

Consider again the two tests A and B described in Exercise2. If a student is chosen at random, what is the probability that the sum of her scores on the two tests will be greater than 200?

In a clinical trial with two treatment groups, the probability of success in one treatment group is 0.5, and the probability of success in the other is 0.6. Suppose that there are five patients in each group. Assume that the outcomes of all patients are independent. Calculate the probability that the first group will have at least as many successes as the second group.

Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the uniform distribution on the interval [0, 1]. Let \({{\bf{Y}}_{\bf{1}}}{\bf{ = min}}\left\{ {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}} \right\}\), \({{\bf{Y}}_{\bf{n}}}{\bf{ = max}}\left\{ {{{\bf{X}}_{\bf{1}}}{\bf{ \ldots }}{{\bf{X}}_{\bf{n}}}} \right\}\)and \({\bf{W = }}{{\bf{Y}}_{\bf{n}}}{\bf{ - }}{{\bf{Y}}_{\bf{1}}}\). Show that each of the random variables \({{\bf{Y}}_{\bf{1}}}{\bf{,}}{{\bf{Y}}_{\bf{n}}}\,\,{\bf{and}}\,\,{\bf{W}}\) has a beta distribution.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.