91影视

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.

Find the expected total prize shared by the players,

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

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

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

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

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

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.

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),

$\begin{array}{r}1鈭�(1鈭�p{)}^{n+1}=E(W)\end{array}$

$1鈭�(1鈭�p{)}^{n+1}=\underset{i=1}{\overset{n+1}{鈭�}}鈥�W_i$

localid="1647500883344" $E\left(W_i\right)=E\left(X\right)$is the expected prize of playerlocalid="1647500893021" $A,$

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

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

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

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.

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

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

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

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

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

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

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

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

From part (b) Calculation,

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

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