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Expert-verified Found in: Page 356 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # 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 $E\left[X\right]=\frac{1-\left(1-p{\right)}^{n+1}}{n+1}$(c) Compute $E\left[X\right]$ by conditioning on whether$A$is a winner, and conclude that $E\left[\left(1+B{\right)}^{-1}\right]=\frac{1-\left(1-p{\right)}^{n+1}}{\left(n+1\right)p}$When $B$ is a binomial random variable with parameters $n$ and $p.$

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

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

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

See the step by step solution

## Step 1: Given Information (Part a)

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

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

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

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,

$E\left(W\right)=P\left(W=1\right)$

$=P\left(\text{At least one winner of the game}\right)$

$=1-P\left(\text{No winner of the game}\right)$

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

## Step 3: Final Answer (Part a)

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

## Step 1: Given Information (Part b)

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

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

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

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 ${i}^{\text{th}}$player.

Each and every player wins independently and has equal probabilities.

From part (a),

$\begin{array}{r}1-\left(1-p{\right)}^{n+1}=E\left(W\right)\end{array}$

$1-\left(1-p{\right)}^{n+1}=\sum _{i=1}^{n+1} {W}_{i}$

localid="1647500883344" $E\left({W}_{i}\right)=E\left(X\right)$ is the expected prize of player localid="1647500893021" $A,$

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

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

## Step 3: Final Answer (Part b)

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

## Step 1: Given Information (Part c)

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

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

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

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 $E\left(X\right)$ by conditioning on whether $A$ is winner.

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

Let $Y=\left\{\begin{array}{l}1 \text{if}A\text{wins}\\ 0 \text{Otherwise}\end{array}\right\$

$\begin{array}{r}E\left(X\right)=E\left[E\left[X\mid Y\right]\right]\end{array}$

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

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

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

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

From part (b) Calculation,

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

## Step 3: Final Answer (Part c)

Hence, the required expression is, $E\left[\left(1+B{\right)}^{-1}\right]=\frac{1-\left(1-p{\right)}^{n+1}}{\left(n+1\right)p}.$ ### Want to see more solutions like these? 