The probability of getting a $4$ in a throw of a die is $\frac{1}{6}$.

The probability of getting other than $4$ i.e. $\overline{4}$ i.e. numbers $1, 2, 3, 5\& 6$ in a single throw of a die is $\frac{5}{6}$.

Total throws done $=5$.

For the event to be completed in $5^{th}$ throw, $4^{th}$ and $5^{th}$ throw must be $4$ . Also, $3^{rd}$ throw must be other than $4$ i.e. $\overline{4}$. Hence, cases for the first two position can be :

Case I : $1^{st}$ throw must be $4$ and $2^{nd}$ throw must not be $4$ i.e. $\overline{4}$

Thus, $P\left(4\stackrel{-}{4}\stackrel{-}{4}44\right)=\frac{1}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}.\frac{1}{6}=\frac{25}{6^5}$.

Case II : $1^{st}$ throw must not be $4$ i.e.  $\overline{4}$ and $2^{nd}$ throw must not be $4$ i.e. $\overline{4}$

Thus, $P\left(\stackrel{-}{4}\stackrel{-}{4}\stackrel{-}{4}44\right)=\frac{5}{6}.\frac{5}{6}.\frac{5}{6}.\frac{1}{6}.\frac{1}{6}=\frac{125}{6^5}$.

Case III : $1^{st}$ throw must not be $4$ i.e. $\overline{4}$ and $2^{nd}$ throw must be $4$

Thus, $P\left(\stackrel{-}{4}4\stackrel{-}{4}44\right)=\frac{5}{6}.\frac{1}{6}.\frac{5}{6}.\frac{1}{6}.\frac{1}{6}=\frac{25}{6^5}$.

Therefore, probability that experiment will end in the fifth throw given that two fours occurred in succession will be

$P\left(4\stackrel{-}{4}\stackrel{-}{4}44\right)+P\left(\stackrel{-}{4}4\stackrel{-}{4}44\right)+P\left(\stackrel{-}{4}\stackrel{-}{4}\stackrel{-}{4}44\right)$

$=\frac{25}{6^5}+\frac{25}{6^5}+\frac{125}{6^5}$

$=\frac{175}{6^5}$.