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Statistics / A First Course in Probability 9 / Chapter 4 / Problem 26STE

A First Course in Probability | 9th Edition | ISBN: 9780321794772 | Authors: Sheldon Ross

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Let ? be the probability that a geometric random variable

Chapter 4, Problem 26STE

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QUESTION:

Problem 26STE

Let α be the probability that a geometric random variable X with parameter p is an even number.

(a) Find α by using the identity

(b) Find α by conditioning on whether X = 1 or X > 1.

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QUESTION:

Problem 26STE

Let α be the probability that a geometric random variable X with parameter p is an even number.

(a) Find α by using the identity

(b) Find α by conditioning on whether X = 1 or X > 1.

ANSWER:

Solution:

Step1 of 2:

Let us suppose that $\alpha{}$ be the probability that a geometric random variable X with parameter p is an even number.

The pmf is

P(X) = $p(1-p{)}^{k-1}$ .

a). To find the $\alpha{}$ by using the identity $\alpha{}=\sum\limits_{i=1}^{\infty{}}P(X=2i).$

Then,

$\alpha{}=\sum\limits_{i=1}^{\infty{}}P(X=2i)$

$=\sum\limits_{\ }^{\ }p{q}^{2i-1}$

$=p\sum\limits_{\ }^{\ }{q}^{2i-2+1}$

$=p\sum\limits_{\ }^{\ }{q}^{2i-2}{q}^{1}$

$=pq\sum\limits_{\ }^{\ }({q}^{2}{)}^{i-1}$

$=pq\ [\ ({q}^{2}{)}^{0}+({q}^{2}{)}^{1}+({q}^{2}{)}^{2}+..+({q}^{n}{)}^{n}]$ (since, ${x}^{0}+{x}^{1}+....+{x}^{n}=\frac{1}{1-{x}^{\ }})$

= $pq[\frac{1}{1-{q}^{2}}]$

$=(1-q).q[\frac{1}{(1-q)(1+q)}]$

$=\frac{1}{1+q}$

Therefore, $\alpha{}=\frac{1}{1+q}$ .

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Let ? be the probability that a geometric random variable

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Solution Found!

Let ? be the probability that a geometric random variable

Questions & Answers

Study Tools You Might Need

Chapter: 10 Problem: 71P

Chapter: 9 Problem: 7.75

Chapter: 14 Problem: 62E

Chapter: 15 Problem: 65

Chapter: 8 Problem: 13

Chapter: 9 Problem: 9.45

Chapter: 5 Problem: 13

Chapter: 5 Problem: 2

Not The Solution You Need? Search for Your Answer Here:

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