It is true that 2/7 of all bulldogs are named Winston. What is the probability that if you ask the name of each bulldog you encounter at a street fair, the fourth time you are given a name other than Winston will be when you ask about the tenth dog you meet? (You may assume that everyone answers your question truthfully and that you will ask about at least ten dogs.)
PROPOSED REPHRASING Suppose that 2/7 of all bulldogs are named Winston. If you start asking bulldog owners the name of their dog one by one, what is the probability that the tenth person you ask will also be the fourth dog whose name isn't "Winston?" (Assume that everyone answers your question truthfully and that you will ask about at least ten dogs.)
"the fourth time you are given a name other than Winston will be when you ask about the tenth dog you meet?"
This will occur precisely when $3$ of the first $9$ bulldogs are not named Winston and when the $10$th bulldog is not named Winston.
As what the name of the $10$th bulldog is independent of the first nine this will be $P(A)*P(B)$ where $A$ = 3 of the first 9 bulldogs are not named Winston and $B$ = the 10th bulldog is not named Winston.
$P(B) = 1 - \frac 27 = \frac 57$ clearly.
Now given $9$ bulldogs there are ${9 \choose 6}$ ways to choose which three of them are not named Winston. No the probability that exact three you choose are not named winston and the exact six you do not choose are named Winston is $(1-\frac 27)^3*(\frac 27)^6$.
So $P(A) = (\frac 57)^3(\frac 27)^6*{9\choose 6}$.
So $P(A)*P(B) = (\frac 57)^3(\frac 27)^6{9\choose 6}\frac 57 = (\frac 57)^4(\frac 27)^6{9\choose 6} = \frac {5^4*2^6*9!}{7^{10}3!6!}\approx 0.01189484746679522353478835237703$