Number of Rolls of Fair Dice to get '6' and '5' A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$

We cannot get both 5 and 6 on the fifth trial, so $P(X=5 \mid Y=5)=0$
$\mathrm{\large E}(X\mid Y=5) = \sum\limits_{x=1}^{4} x\cdot\mathrm{\large P}(X=x\mid Y=5) + \sum\limits_{x=6}^{\infty} x\cdot\mathrm{\large P}(X=x\mid Y=5)$
To unconditionally get a 5 on the fifth trial, we need four not 5 before a 5 on the fifth.$$\mathrm{\large P}(Y=5) = \left(\frac{5}{6}\right)^4 \frac 16$$ 
To get a 6 before getting a 5 on the fifth trial, we have $(x-1)$ rolls that are neither, a 6 on the $x^{th}$, $(3-x)$ rolls that are not 5, then a 5 on the fifth trial. 
$$\mathrm{\large P}(X=x\mid Y=5)|_{x<5} = \dfrac{\mathrm{\large P}(X=x , Y=5)|_{x<5}}{\mathrm{\large P}(Y=5)} \\ = \dfrac{\left(\frac{4}{6}\right)^{x-1}\frac{1}{6}\left(\frac{5}{6}\right)^{3-x}\frac{1}{6}}{\left(\frac{5}{6}\right)^4\frac{1}{6}} = \dfrac{4^{x-1}5^{3-x}}{6^5}\cdot\dfrac{6^5}{5^4} = \dfrac{1}{20} \cdot\dfrac{4^{x}}{5^{x}}$$
To get a 6 after getting a 5 on the fifth trial, we have $4$ rolls that are neither, a 5 on the fifth trial, $(x-6)$ rolls that are not 6, then a 6 on the $x^{th}$ trial. 
$$\mathrm{\large P}(X=x\mid Y=5)|_{x>5} = \dfrac{\mathrm{\large P}(X=x , Y=5)|_{x>5}}{\mathrm{\large P}(Y=5)} \\ = \dfrac{\left(\frac{4}{5}\right)^{4}\frac{1}{6}\left(\frac{5}{6}\right)^{x-6}\frac{1}{6}}{\left(\frac{5}{6}\right)^4\frac{1}{6}} = \dfrac{4^45^{x-6}}{6^x}\cdot\dfrac{6^5}{5^4} = \dfrac{4^4}{5^5}\cdot\dfrac{5^{x-5}}{6^{x-5}}$$
So $$P(X=x\mid Y=5) = \begin{cases} \dfrac{1}{20}\cdot\dfrac{4^{x}}{5^{x}} & 0< x < 5 \\ \dfrac{4^4}{5^5}\cdot\dfrac{5^{x-5}}{6^{x-5}} & x>5 \\ 0 & \text{elsewhere}\end{cases}$$
Putting it together:
$$\mathrm{\large E}(X\mid Y=5) = \sum\limits_{x=1}^{4} x\cdot\mathrm{\large P}(X=x\mid Y=5|_{x<5} + \sum\limits_{x=6}^{\infty} x\cdot\mathrm{\large P}(X=x\mid Y=5|_{x>5} \\ = \dfrac{1}{20}\sum\limits_{x=1}^{4} x\left(\dfrac{4}{5}\right)^x + \dfrac{4^4}{5^5}\sum\limits_{x=6}^{\infty} x \left(\dfrac{5}{6}\right)^{x-5} \\ = \dfrac{14901}{3125} \approx 4.76832 $$
