1. Interpret the event

We roll a fair die until the first time two consecutive fours appear.

For the experiment to end on the 5th throw, we need:

• throws 4 and 5 to be $4,4$,
• and there must be no earlier occurrence of two consecutive fours in throws 1 to 4.

So we want sequences $(x_1,x_2,x_3,x_4,x_5)$ such that: $x_4=4,\quad x_5=4$ and among pairs $(x_1,x_2)$, $(x_2,x_3)$, $(x_3,x_4)$, none is $(4,4)$.

2. Conditions on the first three throws

Since $x_4=4$, to avoid an earlier ending at throw 4, we must have: $x_3\ne 4$ Otherwise $(x_3,x_4)=(4,4)$ would already end the experiment at throw 4.

Also, among throws 1,2,3, we must avoid consecutive fours, so: $(x_1,x_2) \ne (4,4), \quad (x_2,x_3) \ne (4,4)$ But because $x_3\ne 4$, the pair $(x_2,x_3)$ can never be $(4,4)$. So the only remaining restriction is: $x_3\ne 4 \quad \text{and} \quad (x_1,x_2)\ne(4,4)$

3. Count favorable sequences

We count valid choices for $(x_1,x_2,x_3)$.

• $x_3\ne 4$: this gives $5$ choices.
• $(x_1,x_2)$ can be any of $6\times 6=36$ pairs except $(4,4)$: so $35$ choices.

Hence favorable sequences: $35\times 5 = 175$

Since $x_4=4$ and $x_5=4$ are fixed, total favorable full sequences of length 5 are: $175$

Total possible sequences of 5 die throws: $6^5$

Therefore, $P(\text{experiment ends on 5th throw})=\frac{175}{6^5}$

4. Match with options

$\frac{175}{6^5}$ corresponds to Option D.

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.