91Ó°ÊÓ

A fair die is successively rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5

Since we have that $X~Geom(1/6)$, we have that

$E\left(X\right)=6$

$E\left(X\right)=6$

A fair die is successively rolled. Let $X$ and $Y$ denote, respectively, the number of rolls necessary to obtain a $6$ and a$5$

We are given that$Y=1$. That implies that we have acquired five in the main throw. Along these lines, we begin throwing again and all things considered, we will require $6$throws to acquire a $6$. Consequently

$E(Xâˆ\pounds Y=1)=1+E\left(X\right)=7$

$E(Xâˆ\pounds Y=1)=7$

A fair die is successively rolled. Let $X$ and $Y$ denote, respectively, the number of rolls necessary to obtain a $6$ and a $5$.

Assuming we are given that $Y=5$, that truly intends that in fifth throw we have gotten five and that there is no fives inside initial four throws. Utilizing the law of the absolute assumption, we have that

$E_{Y=5}\left(X\right)=\underset{i=1}{\overset{4}{∑}}{E}_{Y=5}(Xâˆ\pounds X=i)P_{Y=5}(X=i)+E_{Y=5}(Xâˆ\pounds X>5)P_{Y=5}(X>5)$

$=\underset{i=1}{\overset{4}{∑}}i{\left(\frac{4}{6}\right)}^{i-1}\frac{1}{6}+(5+6)·{\left(\frac{4}{6}\right)}^4=\frac{161}{54}$