Why the Wording Matters
A geometric waiting-time problem often requires only a few probability calculations, but the main challenge is deciding which trials the words describe. “On the fifth trial,” “at least five trials,” and “within four trials” are different events. An indexing shift of one trial can change the answer even when the probability formula itself is correct.
The distinction between two common geometric random variables matters as well. One records the trial number of the first success; the other records the number of failures before the first success. They describe the same sequence of trials but use different numerical values. This tutorial focuses on translating the wording into events before calculating.
These assumptions are essential to the geometric model. If trial probabilities change, or the trials are dependent, the geometric probability formulas below generally do not apply. No interest-rate or mortality assumptions are involved in these probability examples.
Translate the Phrase Before Using a Formula
First decide whether the problem describes \(T\), the trial on which success occurs, or \(G\), the failures before success. Then translate the phrase into an event. Only after that should you substitute into a probability formula.
| Wording about the first success | Event in terms of \(T\) | Equivalent event in terms of \(G\) |
|---|---|---|
| On trial \(k\) | \(T=k\) | \(G=k-1\) |
| By the end of trial \(n\); within \(n\) trials | \(T\leq n\) | \(G\leq n-1\) |
| Not within the first \(n\) trials | \(T>n\) | \(G\geq n\) |
| At least \(m\) trials are needed | \(T\geq m\) | \(G\geq m-1\) |
| At most \(r\) failures occur before success | \(T\leq r+1\) | \(G\leq r\) |
“Within four trials” includes success on trial 4, so it means \(T\leq4\), not \(T<4\). “At least five trials are needed” means success does not occur on trials 1 through 4, so \(T\geq5\). The word “at least” does not automatically mean a strict inequality; identify the random variable and the quantity being counted.
For a range, translate both endpoints. For example, “the first success occurs on trial 3, 4, or 5” is \(3\leq T\leq5\). In failures-before-success notation, the same event is \(2\leq G\leq4\). The endpoints shift because \(G=T-1\).
Probability Formulas for the Events
To have the first success on trial \(k\), the first \(k-1\) trials must fail and trial \(k\) must succeed. Independence gives the probability \(q^{k-1}p\). This also explains why the exponent is one less than the trial number.
In terms of \(G\), the same sequence has \(g\) failures followed by a success. Thus \(P(G=g)=q^g p\), for \(g=0,1,2,\ldots\). These formulas agree under \(g=k-1\); they are not competing models.
The cumulative and tail events follow by considering whether success has happened by a given trial. The event \(T\leq n\) is the complement of \(T>n\). The latter requires the first \(n\) trials all to be failures, with probability \(q^n\).
The same facts in the failures convention are \(P(G\leq r)=1-q^{r+1}\) and \(P(G>r)=q^{r+1}\), for \(r=0,1,2,\ldots\). The exponent is the number of failures that must occur, not simply the largest value of \(G\).
For an inclusive interval of trial numbers, subtract the probability of waiting beyond the upper endpoint from the probability of waiting beyond the trial immediately before the lower endpoint. Thus, for integers \(1\leq a\leq b\),
This is also obtained by adding the exact probabilities \(q^{k-1}p\) for \(k=a,\ldots,b\). In failures notation, the same interval is \(a-1\leq G\leq b-1\). In practice, translating the event first is safer than trying to remember a formula from a phrase.
Worked Examples
Worked Example: Exact Trial and a Waiting-Time Range
A hypothetical inspection process has independent inspections, with probability \(p=0.30\) that any one inspection detects a specified defect. Let \(T\) be the inspection number of the first detection. Find (a) the probability that the first detection is on inspection 4, and (b) the probability that at least 5 but no more than 8 inspections are needed.
Method and event translation. Here \(p=0.30\), \(q=1-p=0.70\), and the variable counts inspections through the first detection. “On inspection 4” is \(T=4\). “At least 5 but no more than 8” is \(5\leq T\leq8\), including both endpoints.
(a) Exact inspection. The first three inspections must fail and the fourth must detect the defect:
(b) Inclusive range. Use the tail difference for \(5\leq T\leq8\):
Rounded to four decimal places, the answers are \(0.1029\) and \(0.1825\). As a check, adding the exact probabilities for inspections 5 through 8 gives \(0.07203+0.050421+0.0352947+0.02470629=0.18245199\), the same interval probability.
Common mistake avoided: The probability of first detection on inspection 4 is not \(q^4p\). That expression incorrectly requires four failures before the success, which would put the first success on inspection 5.
Worked Example: Adding Several Exact Trial Outcomes
A hypothetical quality-control test succeeds independently with probability \(p=0.40\) on each test. Let \(T\) be the test number of the first success. Find the probability that the first success occurs on test 3, 4, or 5. Give the event in both geometric conventions.
Method and event translation. The specified trials form the event \(3\leq T\leq5\). Since \(G=T-1\), the same event is \(2\leq G\leq4\). Here \(q=1-0.40=0.60\).
Add the three mutually exclusive exact-trial probabilities:
Rounded to four decimal places, the probability is \(0.2822\). A second check uses the tail difference:
Common mistake avoided: Do not interpret “test 3, 4, or 5” as three failures before success. The test number is \(T\), while the number of failures before success is \(G\); the corresponding \(G\)-values are 2, 3, and 4.
Worked Example: “Not Within” Followed by “By” a Deadline
A hypothetical application review is complete on each independent review with probability \(p=0.20\). Let \(T\) be the review number of the first completed application. Find the probability that it is not complete within the first 3 reviews but is complete by the end of review 7.
Method and event translation. “Not complete within the first 3” means the first three reviews are not successes, so \(T>3\). “Complete by the end of review 7” means \(T\leq7\). Together, the event is \(4\leq T\leq7\). Here \(q=1-0.20=0.80\).
The event consists of first success on review 4, 5, 6, or 7. Using a tail difference,
Rounded to four decimal places, the probability is \(0.3023\). In failures notation, the event is \(3\leq G\leq6\): the first three reviews fail, and the first success arrives by review 7.
Common mistake avoided: “Not within the first 3” alone is \(T>3\), not \(T\geq3\). Review 3 is still within the first three, so a success on review 3 must be excluded.
Common Translation Traps
Most errors arise before the arithmetic. Use these checks whenever a problem describes the timing of the first success.
- Check what is counted. “Trial number” uses \(T\), which starts at 1; “failures before success” uses \(G\), which starts at 0.
- Keep endpoints inclusive when the wording is inclusive. “By trial 6” means \(T\leq6\), and “from trial 3 through trial 6” means \(3\leq T\leq6\).
- Count the required failures. Success on trial \(k\) requires \(k-1\) failures, not \(k\). The probability is \(q^{k-1}p\).
- Translate “not within” as a strict tail. Not within the first \(n\) trials means \(T>n\), equivalent to \(T\geq n+1\).
- Use the complement with the right number of trials. “Within \(n\) trials” is \(1-q^n\), since failure to meet the condition requires all \(n\) trials to fail.
Check Your Understanding
For each problem, identify the event in the stated convention before calculating. Probabilities are rounded to four decimal places.
- A hypothetical sensor detects a condition independently with probability \(p=0.25\) per scan. If \(T\) is the scan number of the first detection, find the probability that detection occurs on scan 4. (Answer: \(T=4,\ P(T=4)=0.1055\))
- A hypothetical inspection has independent success probability \(p=0.30\) on each trial. If \(T\) is the trial number of the first success, find the probability of success within the first 3 trials. (Answer: \(T\leq3,\ P(T\leq3)=0.6570\))
- A hypothetical calibration attempt succeeds independently with probability \(p=0.40\). Find the probability that the first success occurs at least on attempt 4 and no later than attempt 6. (Answer: \(4\leq T\leq6,\ P=0.1693\))
- A hypothetical review is completed independently with probability \(p=0.20\) per review. Find the probability that it is not completed within the first 2 reviews but is completed by the end of review 5. (Answer: \(3\leq T\leq5,\ P=0.3123\))
- A hypothetical device passes an independent test with probability \(p=0.35\) each time. Let \(G\) count failures before the first pass. Find the probability of at most 2 failures before the first pass, and express the event in terms of \(T\) as well. (Answer: \(G\leq2\), equivalently \(T\leq3\), with probability \(0.7254\))