<s>[INST] Find the probability of getting two sixes in $5$ throws of a die. 
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial? 

My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.
 [/INST] Condition A: The first two rolls musn't be both 6:
    $$A=1-(\frac{1}{6}\times\frac{1}{6})=\frac{35}{36}$$
Condition B: The third roll mustn't be 6:
    $$B=1-\frac{1}{6}=\frac{5}{6}$$
Condition C: The last two rolls must be 6:
    $$C=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$
Probability of A, B and C being true at the same time:
    $$A\bigcup B\bigcup C=A\times B\times C=\frac{175}{6^5}=\frac{175}{7776}$$
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