91影视

Given in the question that, you plan to play a $5$games. Each game you play is a win with probability $p$. we have to find the expected number of games that you play.

Here, we are going to play $5$games.

Let's consider the first four games separately. From the $5^{th}$game and on we need to play until we lose.

Therefore, the total number of games can be indicated as $4+x$.

Where $X$has geometric distribution with parameter of success $1-p$

Hence, the expected number of games that will be played is

$E(4+X)=4+EX=4+\frac{1}{1-p}$

The expected number of games that will be played is $E\left(4+X\right)=4+\frac{1}{1-P}$

Given in the question that, you plan to play a $5$games. Each game you play is a win with probability $p$. We have to find the expected number of games that you lose.

Consider $Y$as the number of games that we lose.

So, $Y$can be written as $Y=Z+1$, where $Z$indicates the number of games that we lose within first four games.

Hence,$E\left(Y\right)=1+E\left(Z\right)=1+4(1-p)$

The expected number of games that you lose is$E\left(Y\right)=1+4\left(1-P\right)$