logic behind Negative binomial distribution

Question

Is reaching 30th failure before 3rd success equivalent to reaching 0,1 or 2 successes before 30th failure?

that is if Y counts number of successes before k-th failure and \begin{equation} \mathbb{P}(Y=y)\begin{pmatrix} y+k-1\\ y \end{pmatrix} (1-p)^k p^y \end{equation} then is it correct that probability of reaching 30th failure before 3rd sucess is equal to \begin{equation} \sum_{i=0}^2\mathbb{P}(Y=i) \end{equation}

Answer 1

Yes. If we set $k = 30$, then $\mathbb P(Y = 0) + \mathbb P(Y = 1) + \mathbb P(Y = 2)$ is the probability of reaching 30 failures before 3 successes.