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Let X follows a binomial distribution with parameters n, p and Y has a negative binomial distribution with parameters r and p, that is,

\(X \sim Bin\left( {n,p} \right)\)

\(Y \sim NB\left( {r,p} \right)\).

According to the binomial distribution, event \(\left\{ {X \ge r} \right\}\) is all the outcomes that satisfy the following logic 鈥搃n n trials, we get r or more success.

According to the negative binomial distribution, this is equal to 鈥� r-th success happens in the (n-r) th trial or before that. In other words, at least n-r trials are required to get r successes.

\(\begin{aligned}{}P\left\{ {X \ge r} \right\} &= P\left\{ {at\,\,least\,\,r\,\,successes\,\,in\,\,n\,\,trials} \right\}\\ &= P\left\{ {\,{r^{th}}\,\,successes\,\,in\,\,{{\left( {n - r} \right)}^{th}}\,\,trials\,\,or\,\,before} \right\}\\ &= P\left\{ {\,Y \le n - r} \right\}\end{aligned}\)

\(P\left\{ {X \ge r} \right\} = P\left\{ {\,Y \le n - r} \right\}\)

Subtract 1 from both sides,

\(\begin{array}{c}1 - P\left\{ {X \ge r} \right\} = 1 - P\left\{ {\,Y \le n - r} \right\}\\P\left\{ {X < r} \right\} = P\left\{ {\,Y > n - r} \right\}\end{array}\)

Therefore, this concludes that,

\({\rm{P}}\left\{ {{\rm{less}}\,\,{\rm{than}}\,\,{\rm{r}}\,\,{\rm{successes}}\,\,{\rm{in}}\,\,{\rm{n}}\,\,{\rm{trials}}} \right\}{\rm{ = P}}\left\{ {\,{\rm{more}}\,\,{\rm{than}}\,\,{\rm{n - r}}\,\,{\rm{trials}}\,\,{\rm{to}}\,\,{\rm{get}}\,\,{\rm{r}}\,\,{\rm{successes}}} \right\}\).