Why the Convention Matters
A geometric or negative binomial model describes the number of trials or failures needed to achieve a specified number of successes in repeated trials. The probability model can be identical while the random variable is defined differently: one convention counts the successful trial, and the other does not. That difference shifts the support and changes the mean, so recognizing the convention is essential before using a formula.
Assume throughout that trials are independent, each trial has the same success probability \(p\), and \(0<p<1\). Write \(q=1-p\) for the failure probability. The number of successes sought, \(r\), is a positive integer. These assumptions matter: if trial probabilities change or outcomes are dependent, the formulas below generally do not apply.
Read the Definition Before Choosing a Formula
The word “geometric” is used for two closely related variables. When \(r=1\), \(T_1\) counts trials until the first success, while \(K_1\) counts failures before the first success. Their supports begin at 1 and 0, respectively. The negative binomial distribution extends the same distinction to \(r\) successes.
The probability mass functions show precisely where the shift enters. For \(T_r=t\), there must be \(r-1\) successes among the first \(t-1\) trials and a success on trial \(t\). For \(K_r=k\), there must be \(r-1\) successes among the first \(k+r-1\) trials, followed by the \(r\)th success.
The two formulas agree under \(t=k+r\), because then \(t-1=k+r-1\) and \(t-r=k\). Thus \(P(T_r=t)=P(K_r=t-r)\) for \(t\ge r\). In particular, \(T_r\) and \(K_r\) are not the same random variable: they differ by the constant \(r\).
| Variable | What is counted? | Support | Probability at a support value |
|---|---|---|---|
| \(T_r\) | Trials through the \(r\)th success, including that success | \(r,r+1,\ldots\) | \(\binom{t-1}{r-1}p^r q^{t-r}\) |
| \(K_r\) | Failures before the \(r\)th success | \(0,1,\ldots\) | \(\binom{k+r-1}{r-1}p^r q^k\) |
Worked Example: Translating an Exact-Value Probability
A device passes an inspection with probability \(p=0.25\) on each independent inspection. Let \(T_3\) be the inspection number of the third pass, and let \(K_3\) be the number of failures before that third pass. Find the probability that the third pass occurs on inspection 7, expressing the event both ways.
Method. Translate the event first: if the third pass occurs on inspection 7, then there were \(7-3=4\) failures before it. Thus \(\{T_3=7\}\) and \(\{K_3=4\}\) are the same event for this process.
Using \(T_3\), the first 6 inspections must contain exactly 2 passes, and inspection 7 must be a pass. Here \(q=1-0.25=0.75\), so
Using \(K_3\) gives the same probability:
To four decimal places, the probability is \(0.0742\). The two calculations check each other. A common mistake is to use \(\binom{7}{2}\) for \(T_3=7\): the final inspection is already fixed as a pass, so the 2 earlier passes are chosen from the first 6 inspections, not all 7.
Means and Variances Under Both Conventions
For a fixed \(r\), the trial-count and failure-count variables have different means but the same variance. The shift by \(r\) increases the mean by \(r\), while adding a constant does not change variance.
These formulas can be justified by considering the waiting time between successive successes. Let \(W_j\) be the number of trials from just after the \((j-1)\)st success through the \(j\)th success. Each \(W_j\) has the trials-until-first-success geometric distribution, and the waiting times are independent under the independent, identically distributed trial model. For \(n=1,2,\ldots\), \(P(W_j=n)=q^{n-1}p\).
Using the geometric-series identities \(\sum_{n=1}^{\infty}nq^{n-1}=1/(1-q)^2\) and \(\sum_{n=1}^{\infty}n^2q^{n-1}=(1+q)/(1-q)^3\), with \(1-q=p\), gives
Consequently, \(\operatorname{Var}(W_j)=E[W_j^2]-(E[W_j])^2=q/p^2\). Since \(T_r=W_1+\cdots+W_r\), independence gives \(E[T_r]=r/p\) and \(\operatorname{Var}(T_r)=rq/p^2\). Also \(K_r=T_r-r\), so \(E[K_r]=r/p-r=rq/p\), and \(\operatorname{Var}(K_r)=\operatorname{Var}(T_r)=rq/p^2\). This also explains why the two means differ by exactly \(r\), not by a change in the underlying success process.
Worked Example: Calculating a Tail and Both Moments
A sensor independently reports an alert on each check with probability \(p=0.40\). Let \(T_3\) be the check number of the third alert, and \(K_3\) the number of non-alerts before the third alert. Find the means and variances under both conventions, and find \(P(T_3>5)\).
Here \(q=1-0.40=0.60\) and \(r=3\). The moment formulas give
The means differ by 3, as required by \(T_3=K_3+3\), and the variances are equal. To find \(P(T_3>5)\), observe that after 5 checks the third alert has not yet occurred exactly when there have been at most 2 alerts in those 5 checks. The number of alerts in 5 checks has a binomial distribution with 5 trials and success probability 0.40, so
To four decimal places, \(P(T_3>5)=0.6826\). Equivalently, \(T_3>5\) means \(K_3\ge3\), since \(T_3=K_3+3\). The event is not \(K_3>3\): exactly 3 failures before the third alert corresponds to \(T_3=6\), which is included in \(T_3>5\).
Combining Conventions Across Independent Processes
If two random variables are independent, expectations always add, and variances add as well. Independence is required for the variance sum without a covariance term, and for multiplying probabilities of events involving the separate processes. It is not required to add expectations.
The variance identity in this form assumes \(X\) and \(Y\) are independent. Without independence, the correct identity is \(\operatorname{Var}(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)\). The following example applies these rules while keeping the two support conventions distinct.
Worked Example: An Independent Geometric and Negative Binomial Count
Two independent processes each have success probability \(p=0.50\) on every trial. In the first process, \(T\) is the trial of its first success. In the second, \(K\) is the number of failures before its second success. Find \(E[T+K]\), \(\operatorname{Var}(T+K)\), and \(P(T>2,\ K=1)\).
Method. \(T\) uses the trials-until-success convention with \(r=1\), while \(K\) uses the failures-before-success convention with \(r=2\). Here \(q=0.50\).
For \(T\),
For \(K\),
Linearity of expectation gives \(E[T+K]=2+2=4\). Since the processes are independent, their covariance is 0, and
For the probability, \(T>2\) means the first two trials of the first process are failures, so \(P(T>2)=(0.50)^2=0.25\). For the second process, \(K=1\) means exactly one failure before its second success. Among the first \(1+2-1=2\) trials there must be one success and one failure, in either order, followed by a success:
The events involve independent processes, so their joint probability is \(0.25(0.25)=0.0625\). The checked results are \(E[T+K]=4\), \(\operatorname{Var}(T+K)=6\), and \(P(T>2,\ K=1)=0.0625\). A tempting but incorrect calculation is \(2/0.50+2(0.50)/0.50=6\): the first term incorrectly uses \(2/p\) for a first-success trial count. The correct formula is \(E[T]=1/p=2\), not \(2/p\).
A Reliable Exam Procedure
When a problem uses “until,” “before,” “trials,” or “failures,” those words determine the random variable. A short translation step before calculation prevents most support errors.
Decide whether the variable counts trials through the target success or failures before it. Record the target number of successes \(r\).
Use \(r,r+1,\ldots\) for \(T_r\), and \(0,1,\ldots\) for \(K_r\). A proposed value outside that support has probability 0.
Use \(T_r=K_r+r\). For example, \(T_r=t\) corresponds to \(K_r=t-r\), while \(T_r>n\) corresponds to \(K_r>n-r\).
Means under the two conventions must differ by \(r\), and their variances must match. For a tail, check whether the boundary is included by translating the event explicitly.
A common boundary error is to convert a strict inequality without accounting for integer support. For example, \(T_r>n\) is equivalent to \(K_r>n-r\), or, because \(K_r\) is integer-valued, to \(K_r\ge n-r+1\). Writing the event in words—“the \(r\)th success has not occurred by trial \(n\)”—is a useful check.
Careful convention handling is aligned with the SOA Exam P (Probability) syllabus because the same negative binomial process can be described in either form, and a one-unit or \(r\)-unit shift can change an exact probability, a tail event, or an expected value.
Check Your Understanding
For each problem, identify the convention before using a formula. Probabilities are rounded to four decimal places when needed.
- Independent trials have success probability \(0.20\). Let \(T_2\) be the trial of the second success. Find \(P(T_2=6)\). (Answer: \(0.0819\))
- For independent trials with success probability \(0.30\), let \(K_4\) count failures before the fourth success. Find \(E[K_4]\), \(E[T_4]\), and \(\operatorname{Var}(K_4)\). (Answer: \(E[K_4]=9.3333\), \(E[T_4]=13.3333\), \(\operatorname{Var}(K_4)=31.1111\))
- Each trial succeeds independently with probability \(0.40\). Let \(T_2\) be the trial of the second success. Find \(P(T_2\le4)\). (Answer: \(0.5248\))
- Two independent processes have success probability \(0.25\) per trial. Let \(T\) be the first process’s trial of its first success, and let \(K\) be the second process’s failures before its third success. Find \(E[T+K]\) and \(\operatorname{Var}(T+K)\). (Answer: \(E[T+K]=13\), \(\operatorname{Var}(T+K)=48\))
- Two independent processes each have success probability \(0.50\) per trial. Let \(T\) be the first process’s trial of its first success and \(K\) the second process’s failures before its second success. Find \(P(T>3,\ K\le1)\). (Answer: \(0.0625\))