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(D) Let $X_i$ be the outcome of the $i$-th throw. We want the experiment to end at the $5$-th throw,meaning the $4$-th and $5$-th throws must be $4$,a

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

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(D) Let $X_i$ be the outcome of the $i$-th throw. We want the experiment to end at the $5$-th throw,meaning the $4$-th and $5$-th throws must be $4$,and the $3$-rd throw must not be $4$ (otherwise it would have ended at the $4$-th throw). The possible sequences of length $5$ ending in $44$ without ending earlier are: $1$. $4, X_2, X_3, 4, 4$ where $X_2 
eq 4$ and $X_3 
eq 4$. Probability $= \frac{1}{6} 	imes \frac{5}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $2$. $X_1, 4, X_3, 4, 4$ where $X_1 
eq 4$ and $X_3 
eq 4$. Probability $= \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $3$. $X_1, X_2, X_3, 4, 4$ where $X_3 
eq 4$ and the sequence does not contain $44$ in the first $4$ throws. Actually,the condition is that the sequence ends at the $5$-th throw. This means the sequence must be of the form $S_1, S_2, S_3, 4, 4$ where $S_3 
eq 4$ and no $44$ occurs in the first $4$ positions. The valid sequences are: $(4, X_2, X_3, 4, 4)$ where $X_2, X_3 
eq 4$: $\frac{1}{6} 	imes \frac{5}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $(X_1, 4, X_3, 4, 4)$ where $X_1, X_3 
eq 4$: $\frac{5}{6} 	imes \frac{1}{6} 	imes \frac{5}{6} 	imes \frac{1}{6} 	imes \frac{1}{6} = \frac{25}{6^5}$. $(X_1, X_2, X_3, 4, 4)$ where $X_3 
eq 4$ and $(X_1, X_2) 
eq (4, 4)$: There are $5 	imes 5 	imes 5 = 125$ such sequences for the first $3$ positions,but we must exclude $(4, 4, X_3)$ which is $5$ sequences. So $125 - 5 = 120$. Wait,the total probability is $\frac{25+25+125}{6^5} = \frac{175}{6^5}$.

In a random experiment,a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to

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