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Q. 9.16

Expert-verified

A First Course in Probability

Found in: Page 413

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

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Short Answer

A random variable can take on any of n possible values x1, ... , xn with respective probabilities p(xi), i = 1, ... , n. We shall attempt to determine the value of X by asking a series of questions, each of which can be answered “yes” or “no.” For instance, we may ask “Is X = x1?” or “Is X equal to either x1 or x2 or x3?” and so on. What can you say about the average number of such questions that you will need to ask to determine the value of X?

The average number of questions to determine the value of $X$ is $E (X) - \frac{n}{2^{n}}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

We have to find the average number of questions that you will need to ask to determine the value of $X$ .

Step 2: Simplify

Mark that the answer to a particular question is $\frac{1}{2}$ i.e.

$P (Yes) = P (No) = \frac{1}{2}$

which means $P (X = k) = {(\frac{1}{2})}^{k}$ for $k = 1, \dots, n$ . So, the average number of questions is

localid="1648130867761" $E (X) = \sum_{k = 1}^{n} k \cdot {(\frac{1}{2})}^{k}$

To calculate, multiply it by $\frac{1}{2}$ .

localid="1648130836360" $\frac{1}{2} E (X) = \sum_{k = 1}^{n} k \cdot {(\frac{1}{2})}^{k + 1}$

Subtracting that from the expression in $(1)$ , we have that

localid="1648130797034" $\frac{1}{2} E (X) = \sum_{k = 1}^{n} k \cdot {(\frac{1}{2})}^{k + 1} - n . {(\frac{1}{2})}^{n + 1}$

Using the formula localid="1648133543693" $\sum_{k = 1}^{n} {(\frac{1}{2})}^{k} = \frac{1}{2} . \frac{1 - 0 . 5^{n + 1}}{1 - 0.5}$ to obtain that the final answer. That is

$E (X) - \frac{n}{2^{n}}$

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