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How is the probability mass function for the negative binomial distribution derived under the total-trials convention?

To finish on trial x, the first x−1x-1 trials must contain exactly r−1r-1 successes, and the x-th trial must be a success. The probability is calculated as the number of ways to arrange r−1r-1 successes in x−1x-1 trials, multiplied by the probability of that specific sequence.

Conditions

  • Trials are independent.
  • Success probability p is constant.
  • x≥rx \ge r.

Reasoning, step by step

  1. Require the final trial (trial x) to be a success.
  2. Require exactly r−1r-1 successes in the preceding x−1x-1 trials.
  3. Calculate the number of ways to choose positions for r−1r-1 successes among x−1x-1 trials using combinations.
  4. Multiply by the probability of r successes and x−rx-r failures.

Example

The formula is C(x−1,r−1)pr(1−p)x−rC(x-1, r-1)p^r(1-p)^{x-r} for integers x≥rx \ge r.

Common misconceptions

  • Forgetting that the last trial must be a success.
  • Using C(x,r)C(x, r) instead of C(x−1,r−1)C(x-1, r-1).

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