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I was reading on negative binomial from a Statistics textbook and came across this portion on probability relation between binomial and negative binomial. $Y$ refers to the number of trials required to get $r$ successes. Can somebody please explain the relation


Based on binomial distribution, event $\{X \geq r\}$ is the set of outcomes that satisfy "$n$ trials led to $r$ successes or more", which is equivalent to "$r$-th success happened at $n$-th trial or before", which is in turn equivalent to "$n$ trials or less were required to get $r$ successes", and that is it. $$\begin{align*} P\{X \geq r\} &= P\{\mbox{at least r successes in n trials}\}\\ &= P\{\mbox{r-th success in n-th trial or before}\}\\ &= P\{\mbox{n or fewer trials to get r successes}\}\\ &= P\{Y \leq n\} \end{align*}$$

The second relation is the complement of first relation that is: $$\begin{align*} P\{X \geq r\} &= P\{Y \leq n\},\\ 1 - P\{X \geq r\} &= 1 - P\{Y \leq n\},\\ P\{X < r\} &= P\{Y > n\}\\ \end{align*}$$

The second relation means:

$$P\{\mbox{less than r successes in n trials}\}= P\{\mbox{more than n trials to get r successes}\}$$

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  • $\begingroup$ Thank you. What about the second relation? $\endgroup$ – Harry Mar 16 '19 at 10:05
  • $\begingroup$ In the second relation, is it right to say that the P( less than r successes in n trials) $=$ P( more than n trials to get r successes) $\endgroup$ – Harry Mar 16 '19 at 10:15
  • $\begingroup$ (32) is simple the complement of (31) $\endgroup$ – Henry Mar 16 '19 at 13:33

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