Why the Remaining Wait Is a New Waiting-Time Problem
Suppose a process consists of independent trials, each with the same probability of success. A fixed number of successes is required, and some have already occurred. The quantity of interest is then the number of additional trials required, not the total number of trials required from the start. Once the observed successes are accounted for, the remaining wait has a negative binomial distribution with a smaller success target.
For example, if a process requires eight successes and five have already been observed, only three more successes are needed. The number of future trials to obtain those three successes is governed by the future trials’ success probability and independence. The number of failures or successes in the completed history does not add to that future wait.
The model requires a fixed success probability on every trial and independence between trials. The earlier history may specify exactly which trials succeeded, or simply how many successes occurred by a fixed observation time: either way, the future remains independent of that history under this model. If the success probability changes over time, or trials are dependent, the result need not hold. If \(s=r\), the target has already been met and the additional wait is zero; the distribution above applies when \(s<r\).
The Conditional Distribution and Its Mean
Write \(q=1-p\). For \(W\) to equal \(w\), the \(w\)th additional trial must be a success. Among the first \(w-1\) additional trials there must be exactly \(K-1\) successes, so that the \(K\)th success occurs on trial \(w\). There are \(\binom{w-1}{K-1}\) ways to place those earlier successes. Each valid sequence has \(K\) successes and \(w-K\) failures.
The conditioning can be omitted when the history is understood: the formula is the distribution of the additional wait after the specified partial completion. It is not the mass function for the total number of trials from the start; the target in the formula is \(K=r-s\), not \(r\).
One way to justify the mean is to divide the future wait into the time to the first of the \(K\) required successes, the time from that success to the next, and so on. Each stage is a first-success geometric waiting time with success probability \(p\), and hence has mean \(1/p\). Adding the \(K\) stage means gives \(E[W]=K(1/p)=K/p\). This uses the linearity of expectation; independence among the stages is not needed to add their means, though the independent-trial model gives each stage the same geometric distribution.
The completion-index formula follows from counting elapsed and future trials separately. If \(n\) trials have already occurred, completion occurs at trial \(n+W\). Therefore \(E[n+W]=n+E[W]=n+K/p\). This is an expected index, not an expected number of additional trials.
Worked Applications
Worked Example: Three Successes Remain in a Review Process
In a hypothetical review process, each claim independently passes a specified check with probability \(p=0.20\). A reviewer needs five passing claims and has already observed two in 11 completed checks. Let \(W\) be the additional checks needed. Find \(P(W=5)\), the expected additional checks, and the expected index of the check on which the target is reached.
Method and remaining target. The model is independent Bernoulli trials with constant success probability \(p=0.20\), so \(q=1-0.20=0.80\). The target is \(r=5\), and \(s=2\) successes have occurred. Thus \(K=r-s=5-2=3\) successes remain. The observed history has \(n=11\) checks.
Exact additional wait. For \(W=5\), the fifth future check must be the third future success. The first four future checks must contain exactly two successes, giving
Expected additional checks.
Expected completion index. Add the 11 checks already completed to the expected future wait: \(11+15=26\). Thus the expected index of the check that completes the target is 26. As a support check, five additional checks is possible because it is at least the three successes still required.
Common mistake avoided. The mass calculation must use \(\binom{4}{2}\), because the final trial is already fixed as a success. Also, \(6(0.008)(0.64)=0.03072\), not \(0.06144\); the latter incorrectly doubles the product.
Worked Example: Two Successes Remain After Nine Tests
In a hypothetical sensor qualification procedure, each test independently passes with probability \(p=0.40\). Qualification requires six passing tests. After nine tests, four have passed. Find the probability that exactly four additional tests are needed, the expected additional number of tests, and the expected index of the test that qualifies the sensor.
Method and remaining target. Here \(r=6\), \(s=4\), so \(K=6-4=2\) successes remain. The future failure probability is \(q=1-0.40=0.60\), and \(n=9\) tests have elapsed.
Exact additional wait. For \(W=4\), there must be one success among the first three additional tests, followed by a success on the fourth:
Expected additional tests.
Expected completion index. The expected index is \(9+5=14\). Four additional tests is a valid outcome since at least two are required. The probability calculation also accounts for the final test being a pass; choosing the locations of both passes freely among all four tests would not enforce that condition.
Common mistake avoided. The nine completed tests affect the expected completion index, but not the distribution or mean of the additional wait. In particular, the expected additional wait is five tests, not \(14\).
Worked Example: Remaining Wait Measured in Screening Days
In a hypothetical daily screening program, each screening independently detects the condition of interest with probability \(p=0.75\). The program aims to record eight detections. After 10 screening days, five detections have occurred. Find the probability that exactly five more screening days are needed, the expected number of additional days, and the expected day on which the target is reached.
Method and remaining target. There are \(K=8-5=3\) detections still required. Each day is one trial, with \(q=1-0.75=0.25\); the 10 elapsed days give \(n=10\).
Exact additional wait. For \(W=5\), the first four future days must contain two detections, and day five must be a detection:
The probability is rounded to four decimal places in the final reported value. The expected additional days are
Therefore the expected day index of the eighth detection is \(10+4=14\). This calculation treats each screening day as exactly one trial; converting the success probability to a different time unit without changing the trial definition would change the model.
Common mistake avoided. The target for the future wait is three detections, not eight. The expected additional time is \(3/0.75\), and the elapsed 10 days are added only when reporting the expected completion day.
Interpreting the Result and Avoiding Exam Traps
The key operation is to condition on the completed history and subtract the successes already observed from the target. Once \(K=r-s\) is identified, the additional wait has the same form as a negative binomial waiting time for \(K\) successes. This is a direct consequence of the independence of future trials, rather than an assumption that the completed history did not happen.
If the observed history is described as \(n\) trials with \(s\) successes, keep the two time quantities distinct:
| Quantity | Meaning | Result |
|---|---|---|
| \(W\) | Additional trials after the observation point | \(E[W]=K/p\) |
| \(n+W\) | Index of the completion trial from the process start | \(E[n+W]=n+K/p\) |
The conditional mass function is also sensitive to the wording “through and including the success.” If the wait is counted in trials, its final trial must be a success. That is why the number of arrangements is \(\binom{w-1}{K-1}\): choose the \(K-1\) earlier successes among the first \(w-1\) trials, then fix the last trial as success. Using \(\binom{w}{K}\) would incorrectly allow sequences whose final trial is a failure.
Check Your Understanding
For each problem, identify the remaining success target before calculating. Assume independent trials with the stated constant success probability.
- A hypothetical process requires nine successes. Six have occurred, and each future trial succeeds with probability \(0.30\). Give the distribution parameters and expected number of additional trials. (Answer: Negative binomial waiting time with \(K=3\) and \(p=0.30\); \(E[W]=10\) trials.)
- A hypothetical inspection requires four passes; one pass has occurred. Each inspection passes independently with probability \(0.50\). Find the probability of exactly five additional inspections and the expected additional number. (Answer: \(P(W=5)=\binom{4}{2}(0.50)^3(0.50)^2=0.1875\); \(E[W]=6\) inspections.)
- In a hypothetical program, 20 trials have produced seven successes. The target is 10 successes, and each future trial succeeds with probability \(0.25\). Find the expected additional trials and the expected completion-trial index. (Answer: \(K=3\), \(E[W]=3/0.25=12\), and the expected completion index is \(20+12=32\).)
- A hypothetical device test requires seven successes. Five have occurred, and each future test succeeds independently with probability \(0.60\). Find \(P(W=3)\) and \(E[W]\). (Answer: \(P(W=3)=\binom{2}{1}(0.60)^2(0.40)=0.2880\); \(E[W]=2/0.60=3.3333\) tests.)
- A hypothetical review requires 12 approvals. Nine approvals have been recorded after 14 trials, and the approval probability on each future trial is \(0.20\). Find the probability of exactly five additional trials and the expected completion-trial index. (Answer: \(K=3\); \(P(W=5)=\binom{4}{2}(0.20)^3(0.80)^2=0.03072\); the expected completion index is \(14+3/0.20=29\).)