I am not quite understanding the solution to this problem.
You are working on a difficult passage from a new piece you are learning on the piano. You wish to play it correctly 4 times before calling it a day. If you have a probability of 2/3 of playing it correctly on every attempt, and the attempts are independent, what is the probability that you attempt to play it at least 8 times?
Solution:
Why it (n-1) nCr 3 and not n nCr 4 ? From my understanding there are 4 'successes' for each n practice sessions.
You wish to find the probability of playing correctly $3$ times among the first $n-1$ tries, then correctly on try $n$ (and you don't care what happens after that), for some $n$ between $4$ and $7$
$$\begin{align}\mathsf P(X<8)~=~& \sum_{n=4}^7\mathsf P(X=n) \\~=~&\sum_{n=4}^7\binom {n-1}3(\tfrac 23)^{3+1}(\tfrac 13)^{n-4}\end{align}$$
PS: You are actually dealing with a "Negative binomial distribution", which is rather different from a Binomial distribution.