/*! 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. 4.31 A jar contains m+n chips, numbe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A jar contains m+nchips, numbered 1,2,…,n+m. A set of size nis drawn. If we let X denote the number of chips drawn having numbers that exceed each of the numbers of those remaining, compute the probability mass function of X.

Short Answer

Expert verified

P(X=k)=n+m-k-1n-kn+mn

Step by step solution

01

Given information

A jar contains m+nchips, numbered 1,2,…,n+m. A set of size n is drawn.

02

Explanation

Observe that X∈{0,…,n}. Take any k∈{0,…,n}. Let's calculate P(X=k). Observe that there are n+mnof all possible combinations of taken chips. If X=k, that means that we have taken klargest number, have not taken (k+1)stlargest number and all other remaining n-knumbers out of remaining n+m-k-1 numbers have been taken freely.

03

Final answer

The probability mass function is

P(X=k)=n+m-k-1n-kn+mn

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

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. What are the possible values of X?

For a hypergeometric random variable, determine

P{X=k+1}/P{X=k}

From a set of n elements, a nonempty subset is chosen at random in the sense that all of the nonempty subsets are equally likely to be selected. Let X denote the number of elements in the chosen subset. Using the identities given in Theoretical Exercise 12of Chapter1, show that

E[X]=n2−12n−1

Var(X)=n⋅22n−2−n(n+1)2n−22n−12

Show also that for n large,

Var(X)~n4

in the sense that the ratio Var(X) ton/4approaches 1as n approaches q. Compare this formula with the limiting form of Var(Y) when P{Y =i}=1/n,i=1,...,n.

Consider ncoins, each of which independently comes up heads with probability p. Suppose that nis large and pis small, and let λ=np. Suppose that all ncoins are tossed; if at least one comes up heads, the experiment ends; if not, we again toss all coins, and so on. That is, we stop the first time that at least one of the ncoins come up heads. Let Xdenote the total number of heads that appear. Which of the following reasonings concerned with approximating P{X=1}is correct (in all cases, Yis a Poisson random variable with parameter λ)?

(a) Because the total number of heads that occur when all ncoins are rolled is approximately a Poisson random variable with parameter λ,

P{X=1}≈P{Y=1}=λe-λ

(b) Because the total number of heads that occur when all ncoins are rolled is approximately a Poisson random variable with parameter λ, and because we stop only when this number is positive,

P{X=1}≈P{Y=1∣Y>0}=λe-λ1-e-λ

(c) Because at least one coin comes up heads, Xwill equal 1 if none of the other n-1coins come up heads. Because the number of heads resulting from these n-1coins is approximately Poisson with mean (n-1)p≈λ,

P{X=1}≈P{Y=0}=e-λ

The random variable X is said to have the Yule-Simons distribution if

P{X=n}=4n(n+1)(n+2),n≥1

(a) Show that the preceding is actually a probability mass function. That is, show that∑n=1∞P{X=n}=1

(b) Show that E[X] = 2.

(c) Show that E[X2] = q

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.