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I am having trouble putting together the way these distributions work. It doesn't matter whether we speak of the support space in terms of number of trials or failures.

Specifically, what variable is it we have control over?

I ask this because, say, we want to find out the probability that I achieve $5$ successes within $10$ trials compared to obtaining $5$ successes within $30$ trials with probability of success being $0.3$ and of failure $0.7$. Numerically I personally would've thought that the probability would have gone up with the larger number of trials, but instead it went down.

Do I control only the number of trials that occur and then whatever the different amounts of successes that occur from said event the i take those as is or do can i also state the amount of successes that I want within a certain range?

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    $\begingroup$ The probability of at least $5$ successes goes up. The probability of exactly $5$ successes goes down. Neither has much to do with the negative binomial, where we specify the number $k$ of successes and are interested in the number of trials until the $k$-th success (or, in an alternate but related definition of the negative binomial, the number of failures before the $k$-th success). We have control over (specify) the $5$, and have no control over the number of trials. By way of contrast, in the binomial we specify the number of trials. $\endgroup$ – André Nicolas Aug 29 '15 at 15:52
  • $\begingroup$ I think the negative binomial makes sense, in regards to how i wrote the question if i phrased it like "given we have 5 successes, what is the probabiity this occured in 10 trials?" So then in this case it isn't the number of successes, but the number of trials it took for those successes $\endgroup$ – dc3rd Aug 29 '15 at 16:04
  • $\begingroup$ It is clearer to say on the $10$-th trial. For $n$ large enough, the probability that the fifth success occurs on the $n$-th trial is decreasing, for we are very likely to have reached the fifth success earlier. So typically the probability the fifth success occurs on the $n$-th trial increases for a while, then decreases. $\endgroup$ – André Nicolas Aug 29 '15 at 16:05
  • $\begingroup$ Got it because within 10 trials would refer to the cumulative aspect, but specifying the 10th trial would make it exact and the use of the mass function. $\endgroup$ – dc3rd Aug 29 '15 at 16:08
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Extended Comment:

I think it would be helpful for you to use more specific language: (a) When you say "the probability that I achieve 5 successes within 10 trials," you need to say 'exactly 5' or 'at least 5'. Even if it is clear to you what you mean, it is not necessarily clear to others. (b) Also, you need to distinguish between what you 'control' for the experiment and what you 'specify' for your event.

$Binomial.$ You control the number of trials $n = 10$. And you control the success probability $\theta = 0.3.$ Then $X_{10} \sim Binom(n, \theta)$ records the number of Successes obtained. You may specify the event $\{X = 5\}$ or $\{X \ge 5\}$.

Then $P(X_{10} = 5)= 0.1029$ and $P(X_{10} \ge 5) = 0.1503$, according to the following computations in R:

 dbinom(5, 10, .3);  sum(dbinom(5:10, 10, .3))
 ## 0.1029193
 ## 0.1502683

For $n = 30,$ you have $P(X_{30} = 5)= 0.0464 < 0.1029$ (as you observed) and $P(X_{30} \ge 5) = 0.9698 > 0.1503$ (as @AndreNicolas commented):

 dbinom(5, 30, .3);  sum(dbinom(5:30, 30, .3))
 ## 0.04643981
 ## 0.969845

I think your problem is more directly handled in terms of binomial probabilities than negative binomial (or geometric) ones. You don't care about the arrangement of successes and failures along the sequence of 10 or 30 trials. In particular, you don't care whether the last trial is a success.

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  • $\begingroup$ @AndreNichoas: For what it's worth, the CDF for neg binom in R, counts numbers of failures before 5th success. For reasons I cannot immediately explain intuitively, pnbinom(10-5, 5, .3) returns 0.1502683 and pnbinom(30-5, 5, .3) returns 0.969845, numbers that appear in the binomial output of my Comment-in-Answer format. Seems equivalent series are being summed. $\endgroup$ – BruceET Aug 30 '15 at 0:30
  • $\begingroup$ Thanks for the added clarification @Bruce_Trumbo. I asked it in terms of negative binomial and geometric because it was the fact that i wws confused over what we control vs what we specify. In terms of the negative binomial, we "control" the number of successes and "specify" the number of trials since it is that which we are curious about in that environment $\endgroup$ – dc3rd Aug 30 '15 at 21:07

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