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Step 1: Understand the Event
We roll a fair die repeatedly until we get two consecutive fours (4–4). We want the probability that this happens for the first time on the fifth throw, meaning the 4th and 5th throws must both be fours, and no two consecutive fours should appear in the first three throws.
Step 2: Identify Conditions for the First Three Throws
For the experiment to end on the 5th throw with two consecutive fours, the sequence of the first three throws must:
Not contain consecutive fours (so no “4–4” in those first three throws).
Not end in a four (otherwise the 4th throw being four would already produce consecutive fours too early).
Step 3: Count Valid Sequences for the First Three Throws
Each throw has 6 possible outcomes (1 through 6). Therefore, there are a total of $6^3 = 216$ possible outcomes for the first three throws.
Let $A_n$ = number of valid sequences of length $n$ that do not contain consecutive fours and end in a non-four.
Let $B_n$ = number of valid sequences of length $n$ that do not contain consecutive fours and end in 4.
These satisfy the following recurrences:
$A_{n+1} = 5\,A_n + 5\,B_n,$ because from any valid sequence of length $n$, we can choose one of the 5 non-four outcomes for the next throw to end in non-four (avoiding “4–4”).
$B_{n+1} = A_n,$ because to end in a 4 at step $n+1$, the $n$th throw must not be a 4.
Initial values for $n=1$:
$A_1 = 5$ (the first throw can be any of 1,2,3,5,6),
$B_1 = 1$ (the first throw is a single 4).
Compute up to $n=3$:
$A_2 = 5\,A_1 + 5\,B_1 = 5\times 5 + 5\times 1 = 25 + 5 = 30,$
$B_2 = A_1 = 5.$
$A_3 = 5\,A_2 + 5\,B_2 = 5\times 30 + 5\times 5 = 150 + 25 = 175,$
$B_3 = A_2 = 30.$
Thus, the number of sequences of length 3 with no consecutive fours and ending in a non-four is $A_3 = 175.$
Step 4: Include the Last Two Throws
Given one such valid sequence of length 3 (probability $175 / 6^3$), the 4th and 5th throws must both be fours. Each has probability $1/6$ independently. Therefore, the probability of the 4th throw being 4 and the 5th throw being 4 is $1/6 \times 1/6 = 1/6^2.$
Step 5: Combine Probabilities
The overall probability is:
$ \displaystyle \left(\frac{175}{6^3}\right) \times \left(\frac{1}{6^2}\right)
=\frac{175}{6^5} .
$
This matches the correct answer.
