Why Geometric Variables Need a Counting Convention
A waiting-time problem asks when the first success occurs in a sequence of trials. The answer depends on what is being counted. One convention counts the trial on which success occurs; the other counts only the failures before that success. The two descriptions refer to the same sequence of outcomes, but their numerical values differ by one.
For example, if the first success occurs on trial 4, then there have been 3 failures before it. A geometric random variable can therefore take the value 4 under the trials convention or 3 under the preceding-failures convention. Correctly identifying the convention is essential before selecting a support or calculating a probability.
These assumptions are the Bernoulli-trial conditions: the trials are independent and have a constant success probability. If the success probability changes from trial to trial, or the trials are dependent, these geometric formulas do not generally apply. The model also assumes that trials continue until a success occurs; in a finite observation window, the first success might not be observed.
The Two Supports and Their Relationship
The support is the set of values a random variable can take. Under the trials convention, the first success cannot occur before trial 1, so \(T\) takes positive integer values. Under the preceding-failures convention, success on the first trial means zero failures came first, so \(F\) can equal zero.
| Convention | Variable | Meaning of value \(k\) | Support |
|---|---|---|---|
| Trials through first success | \(T\) | First success occurs on trial \(k\) | \(1,2,3,\ldots\) |
| Failures before first success | \(F\) | Exactly \(k\) failures occur before the first success | \(0,1,2,3,\ldots\) |
The translation is always a one-unit shift. If the first success occurs on trial \(t\), exactly \(t-1\) failures preceded it. Conversely, if \(f\) failures occur before the first success, the success is on trial \(f+1\).
This relationship also gives a direct translation of events. “The first success is on trial 6” and “there are 5 failures before the first success” describe the same outcome sequence. They do not describe different waiting times; they report the same waiting time using different counts.
Exact Probabilities Under Each Convention
Let \(q=1-p\) be the probability of failure on any one trial. For the first success to occur on trial \(t\), the first \(t-1\) trials must be failures and trial \(t\) must be a success. Independence means the probability of that ordered sequence is the product of its trial probabilities.
For exactly \(f\) failures before the first success, the sequence consists of \(f\) failures followed by one success. Therefore:
The exponents differ because \(T=t\) includes the success trial, while \(F=f\) counts only the failures before that trial. The formulas agree whenever the values are translated correctly: putting \(t=f+1\) in the trials formula gives \(q^{(f+1)-1}p=q^fp\), which is the preceding-failures formula.
The endpoint cases are different from the usual nondegenerate geometric model. If \(p=1\), success occurs immediately, so \(T=1\) and \(F=0\) with probability 1. If \(p=0\), no success ever occurs, so there is no finite first-success time; neither finite-valued geometric variable describes the outcome. The formulas above are used for \(0<p<1\).
Worked Examples: Translating the Convention Before Calculating
Worked Example: A Sensor’s First Positive Reading
A hypothetical sensor is checked repeatedly. Each check independently gives a positive reading with probability \(p=0.30\), and this probability is constant from check to check. Let \(T\) be the check number of the first positive reading. Find the probability that the first positive reading is on check 4, and express the same outcome using the preceding-failures convention.
Method and model. The checks are independent Bernoulli trials with success probability \(p=0.30\) and failure probability \(1-p=0.70\). The variable \(T\) counts trials through the first success, so its support is \(1,2,3,\ldots\).
Translate the wording. A first positive reading on check 4 means checks 1, 2, and 3 are negative, followed by a positive reading on check 4. Thus \(T=4\), or equivalently \(F=4-1=3\).
Calculate. Using the trials convention:
Using the failures convention gives the same result:
The probability is \(0.1029\), or 10.29%. The check is that both descriptions specify three failures followed by one success, so their probabilities match. The common mistake this avoids is using an exponent of 4: the success occurs on check 4, but only three failures precede it.
Worked Example: Denials Before an Approval
In a hypothetical review process, each application is independently denied with probability \(0.55\) and approved with probability \(p=0.45\). Let \(F\) count denials before the first approval, and let \(T\) count applications reviewed through and including the first approval. Find the probability of exactly two denials before the first approval, and give the corresponding value of \(T\).
Method and model. This is a geometric model with independent trials and constant approval probability \(p=0.45\); the failure probability is \(q=0.55\). The support of \(F\) is \(0,1,2,\ldots\), and the support of \(T\) is \(1,2,3,\ldots\).
Translate the count. Exactly two denials before the first approval means \(F=2\). The approval is therefore the third application reviewed, so \(T=F+1=3\).
Calculate using \(F\). The ordered sequence is denial, denial, approval:
Rounded to four decimal places, the probability is \(0.1361\). As a check, \(P(T=3)=(0.55)^{3-1}(0.45)=0.136125\), the same value. A common mistake is to interpret “two denials before approval” as approval on application 2. That would allow only one denial; the correct application number is 3.
Worked Example: Interpreting a First Success on Trial 7
A hypothetical equipment test succeeds on each attempt with probability \(p=0.20\), independently across attempts. Let \(F\) be the number of failed attempts before the first success. Find the probability that the first success occurs on attempt 7, and state the corresponding value of \(F\).
Method and model. The attempts satisfy the geometric assumptions with success probability \(0.20\) and failure probability \(0.80\). Since \(F\) counts only failures before success, its support begins at zero.
Translate to the specified variable. A success on attempt 7 follows six failed attempts, so \(T=7\) and \(F=T-1=6\). The event can be written as \(F=6\).
Calculate. The first six attempts must fail and the seventh must succeed:
Rounded to four decimal places, the probability is \(0.0524\). The equivalent calculation is \(P(T=7)=(0.80)^{7-1}(0.20)\), which produces the same value. The common mistake this avoids is setting \(F=7\): the seventh attempt is the successful attempt and is not counted among the failures.
A Reliable Translation Check
Before calculating, identify what the random variable counts and then map the wording to that count. A quick sequence sketch can make the distinction visible: write one “failure” for each unsuccessful trial, then write the success. If the success is in position \(t\), the sketch has \(t-1\) failures. If the sketch has \(f\) failures before success, its final success is in position \(f+1\).
Determine whether the variable counts trials through success or failures before success.
A trials count starts at 1; a preceding-failures count starts at 0. A value outside that support cannot describe the variable.
Use \(F=T-1\) or \(T=F+1\), then apply the exact-probability formula with the number of failures as the exponent.
The most common exam trap is an off-by-one error in the exponent or the stated value. For instance, if success occurs on trial 5, the incorrect calculation \(q^5p\) counts five failures followed by success, placing success on trial 6. The correction is \(q^{5-1}p=q^4p\). Under the failures convention, the same event is \(F=4\), and the formula \(q^4p\) applies directly.
Check Your Understanding
For each problem, identify the relevant convention before using the geometric model. Assume the stated trials are independent with a constant success probability.
- A hypothetical inspection has success probability \(0.25\) on each inspection. Let \(T\) be the inspection number of the first success. Translate \(T=4\) into a value of \(F\), the number of prior failures, and calculate the probability. (Answer: \(F=3;\ P(T=4)=0.1055\), to four decimals.)
- In a hypothetical approval process, the approval probability is \(0.60\) per application. Find the probability of exactly two failures before the first approval, and state the application number of that approval. (Answer: \(T=3;\ P(F=2)=(0.40)^2(0.60)=0.0960\).)
- A test has success probability \(0.10\) on each attempt. If \(T\) counts attempts through the first success, translate \(T=8\) into the preceding-failures convention and calculate the probability. (Answer: \(F=7;\ P(T=8)=(0.90)^7(0.10)=0.0478\), to four decimals.)
- A hypothetical device passes its first check with probability \(0.37\). Let \(F\) count failures before the first pass. State the support value for an immediate pass and calculate its probability. (Answer: \(F=0;\ P(F=0)=0.37\).)
- A repeated screening procedure has success probability \(0.20\) per screening. A report says that the first success occurs after four failures. State the value of \(T\), write the exact-probability expression, and evaluate it. (Answer: \(T=5;\ P(F=4)=P(T=5)=(0.80)^4(0.20)=0.0819\), to four decimals.)