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A First Course in Probability

Found in: Page 356

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

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Short Answer

There are $n + 1$ participants in a game. Each person independently is a winner with probability $p$ . The winners share a total prize of 1 unit. (For instance, if 4 people win, then each of them receives $\frac{1}{4}$ whereas if there are no winners, then none of the participants receives anything.) Let $A$ denote a specified one of the players, and let $X$ denote the amount that is received by $A .$
(a) Compute the expected total prize shared by the players.
(b) Argue that $E [X] = \frac{1 - (1 - p)^{n + 1}}{n + 1}$
(c) Compute $E [X]$ by conditioning on whether $A$ is a winner, and conclude that $E [(1 + B)^{- 1}] = \frac{1 - (1 - p)^{n + 1}}{(n + 1) p}$
When $B$ is a binomial random variable with parameters $n$ and $p .$

a) The expected total prize shared by the player is $E [W] = 1 - (1 - p)^{n + 1}$

b) The required expression is, $E (X) = \frac{1 - (1 - p)^{n + 1}}{(n + 1)}$

c) Computing $E [X]$ by conditioning on whether $A$ is a winner, and conclude that $E [(1 + B)^{- 1}] = \frac{1 - (1 - p)^{n + 1}}{(n + 1) p}]$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information (Part a)

The probability of each person winning the game is, $p .$

The number of players in the game is, $(n + 1) .$

The probability of the no winner of the game is, $(1 - p)^{n + 1}$

Let $W = 0$ denote the no winner in the game

Let $W = 1$ denote the at least one winner in the game.

Step 2: Explanation (Part a)

Find the expected total prize shared by the players,

$E (W) = P (W = 1)$

$= P (At least one winner of the game)$

$= 1 - P (No winner of the game)$

$= 1 - (1 - p)^{n + 1}$

Step 3: Final Answer (Part a)

Hence, the required value is $1 - (1 - p)^{n + 1} .$

Step 1: Given Information (Part b)

The probability of each person wins the game is, $p .$

The number of players in the game is, $(n + 1)$ .

The probability of the no winner of the game is, $(1 - p)^{n + 1}$

Let $W = 0$ denote the no winner in the game

Let $W = 1$ denote the at least one winner in the game.

Step 2: Explanation (Part b)

Let $W_{i}$ denote the prize of the $i^{th}$ player.

Each and every player wins independently and has equal probabilities.

From part (a),

$\begin{aligned} 1 - (1 - p)^{n + 1} = E (W) \end{aligned}$

$1 - (1 - p)^{n + 1} = \sum_{i = 1}^{n + 1} W_{i}$

localid="1647500883344" $E (W_{i}) = E (X)$ is the expected prize of player localid="1647500893021" $A,$

localid="1647500947723" $1 - (1 - p)^{n + 1} = (n + 1) E (X)$

localid="1647500955566" $\frac{1 - (1 - p)^{n + 1}}{(n + 1)} = E (X)$

Step 3: Final Answer (Part b)

Hence, the required expression is, $E (X) = \frac{1 - (1 - p)^{n + 1}}{(n + 1)} .$

Step 1: Given Information (Part c)

The probability of each person winning the game is, $p .$

The number of players in the game is, $(n + 1) .$

The probability of the no winner of the game is, $(1 - p)^{n + 1}$

Let $W = 0$ denote the no winner in the game

Let $W = 1$ denote the at least one winner in the game.

Step 2: Explanation (Part c)

Find the $E (X)$ by conditioning on whether $A$ is winner.

Let $B$ follows a Binomial random variable with parameter $(n, p)$ .

Let $Y = \{\begin{cases} 1 if A wins \\ 0 Otherwise \end{cases}$

$\begin{aligned} E (X) = E [E [X ∣ Y]] \end{aligned}$

$\begin{array}{r} E (X) = E [X ∣ Y = 0] P (Y = 0) + E [X ∣ Y = 1] P (Y = 1) \end{array}$

$\begin{array}{r} E (X) = 0 + E [(1 + B)^{- 1}] (p) \end{array}$

$E (X) = E [(1 + B)^{- 1}] (p)$

$\frac{E (X)}{p} = E [(1 + B)^{- 1}]$

From part (b) Calculation,

$\frac{1 - (1 - p)^{n + 1}}{(n + 1) p} = E [(1 + B)^{- 1}]$

Step 3: Final Answer (Part c)

Hence, the required expression is, $E [(1 + B)^{- 1}] = \frac{1 - (1 - p)^{n + 1}}{(n + 1) p} .$

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