In my statistics book $B$ is defined as: $$B(x; \ n, \ p) = \sum_{y = 0}^{x}b(y; \ n, \ p) = \sum_{y = 0}^{x}{n \choose y}p^{y}(1 - p)^{n - y}$$
I want to show that: $$B(x; \ n, \ 1 - p) = 1 - B(n - x - 1; \ n, \ p)$$
The hint that I got was that at most $x$ successes is equivalent to at least $(n - x)$ failures in the binomial probability distribution. I tried to rewrite it but I get stuck in nowhere. Appreciate any help, cheers.
\begin{align} 1 - B(n - x - 1; \ n, \ p) = & 1 - \sum_{y = n-x-1}^{n}{n \choose y}p^{y}(1 - p)^{n - y} \\ = & \sum_{y = 0}^{n}{n \choose y}p^{y}(1 - p)^{n - y} - \sum_{y = n-x-1}^{n}{n \choose y}p^{y}(1 - p)^{n - y} \\ = & \sum_{y = n-x}^{n}{n \choose y}p^{y}(1 - p)^{n - y} \\ = & \binom{n}{n-x}p^{n-x}(1-p)^{x} + \binom{n}{n-x+1}p^{n-x+1}(1-p)^{x-1} + ... + \binom{n}{n}p^{n}(1-p)^{0} \\=& \binom{n}{n-(n-x)}p^{n-x}(1-p)^{x} + \binom{n}{n-(n-x+1)}p^{n-x+1}(1-p)^{x-1} + ... + \binom{n}{n-(n)}p^{n}(1-p)^{0} \\=& \binom{n}{x}p^{n-x}(1-p)^{x} + \binom{n}{x-1}p^{n-x+1}(1-p)^{x-1} + ... + \binom{n}{0}p^{n}(1-p)^{0} \\ =& \sum_{y = 0}^{x}{n \choose y}(1-p)^{y}p^{n - y} \\ =& B(x; \ n, \ 1 - p) \end{align}