First-Success Geometric Problems With Changing Wording

Turning Scenario Wording Into a First-Success Event

A first-success problem is often straightforward once the event is stated precisely. The difficulty is that ordinary phrases such as “needs at least five attempts,” “passes by the fourth test,” and “the first defect is found after the second inspection” refer to different boundaries. A one-trial shift can change the probability substantially.

The first step is to define what counts as a success in the scenario. Then define the trial number on which the first success occurs. The success need not be a favorable outcome in everyday language: it is simply the outcome being tracked, such as finding a defect or obtaining a positive test result.

Definition: Suppose a sequence of trials is independent, and each trial has the same probability $p$ of success, where $0<p\leq 1$. Let $q=1-p$, and let $T$ be the trial number of the first success. Then $T$ has a geometric distribution on $1,2,3,\ldots$, with $$ P(T=k)=q^{k-1}p,\qquad k=1,2,3,\ldots. $$ The event $T=k$ requires failures on the first $k-1$ trials and a success on trial $k$.

This tutorial uses the “trials until first success” convention, so the smallest possible value of $T$ is 1. Before calculating, verify that the model’s assumptions fit the scenario: the trials must be independent and the success probability must be the same on each trial. If either condition does not hold, the geometric formulas below do not directly apply.

A Translation Guide for Common Phrases

The geometric tail is especially useful for translating wording. For any positive integer $n$, $T>n$ means the first $n$ trials were all failures. Thus $P(T>n)=q^n$. The distinction between $T>n$ and $T\geq n$ matters: the latter requires only the first $n-1$ trials to be failures.

Wording or event	Event in terms of $T$	Probability
First success on trial $k$	$T=k$	$q^{k-1}p$
Success by trial $n$, inclusive	$T\leq n$	$1-q^n$
No success in the first $n$ trials	$T>n$	$q^n$
At least $n$ trials are needed	$T\geq n$	$q^{n-1}$
First success from trial $a$ through trial $b$, inclusive	$a\leq T\leq b$	$q^{a-1}-q^b$
First success after trial $a$, but by trial $b$	$a<T\leq b$	$q^a-q^b$

The interval formulas are differences of tail probabilities. For instance, $a<T\leq b$ includes outcomes where the first success has not occurred by trial $a$, but has occurred by trial $b$. It is therefore the event $T>a$ with the event $T>b$ removed.

Why the Tail Difference Gives an Interval Probability

A short derivation explains the interval translation and helps prevent endpoint errors. Assume the independent, constant-$p$ model above, and take integers $0\leq a<b$. The event $T>a$ means the first $a$ trials fail; the event $T>b$ means the first $b$ trials fail. Every outcome with $T>b$ is also in $T>a$. Removing those outcomes leaves exactly $a<T\leq b$.

\begin{aligned} P(a<T\leq b) &=P(T>a)-P(T>b)\\ &=q^a-q^b\\ &=q^a(1-q^{b-a}). \end{aligned}

The last expression has a direct interpretation: the first $a$ trials must fail, and then at least one success must occur in the next $b-a$ trials. This argument also shows why “after $a$, by $b$” does not mean $a\leq T\leq b$: the success on trial $a$ is excluded.

Worked Geometric Problems

Worked Example: Finding a Defect During an Inspection Sequence

A hypothetical inspection process checks items one at a time. For each item, the chance that inspection reveals a specified defect is 0.20, independently of other inspections. Let $T$ be the inspection number on which the defect is first revealed. Find (a) the probability the first defect is revealed on inspection 3, and (b) the probability it is first revealed after inspection 4 but by inspection 7.

Model and method. Each inspection is one trial, and revealing the defect is success. The model is geometric with $p=0.20$ and $q=1-p=0.80$. There is no interest or mortality assumption in this inspection example. Use the exact-trial formula for part (a) and a difference of tails for part (b).

(a) Exactly inspection 3. The first two inspections must not reveal the defect, and the third must reveal it:

$$ P(T=3)=q^2p=(0.80)^2(0.20)=0.64(0.20)=0.128. $$

Rounded to four decimal places, $P(T=3)=0.1280$.

(b) After inspection 4 and by inspection 7. This wording means $4<T\leq 7$, not $4\leq T\leq 7$. Subtract the probability of no defect revealed in the first seven inspections from the probability of no defect revealed in the first four:

\begin{aligned} P(4<T\leq 7) &=P(T>4)-P(T>7)\\ &=(0.80)^4-(0.80)^7\\ &=0.4096-0.2097152\\ &=0.1998848. \end{aligned}

Rounded to four decimal places, the probability is $0.1999$. As a check, the factored form gives $(0.80)^4[1-(0.80)^3]=0.4096(0.488)=0.1998848$, the same result.

Common mistake avoided: Including inspection 4 would answer a different event. “After inspection 4” excludes a first finding on inspection 4, so the event begins at inspection 5.

Worked Example: Translating an Attempt Requirement

In a hypothetical equipment-calibration procedure, each attempt independently succeeds with probability 0.40. Let $T$ be the attempt number of the first successful calibration. Find the probability that at least five attempts are needed, and the probability that calibration succeeds by attempt 3.

Model and method. Success means successful calibration, with $p=0.40$ and $q=0.60$, so $T$ is geometric on the positive integers. No interest or mortality assumption is involved. Translate the two phrases separately: “at least five attempts are needed” means $T\geq 5$, while “by attempt 3” means $T\leq 3$.

At least five attempts. To need five or more attempts, the first four must fail. A success on attempt 5 is allowed, so this event is not $T>5$:

P(T\geq 5)=P(T>4)=q^4=(0.60)^4=0.1296.

Success by attempt 3. Use the complement of failing all three attempts:

\begin{aligned} P(T\leq 3) &=1-P(T>3)\\ &=1-q^3\\ &=1-(0.60)^3\\ &=1-0.216\\ &=0.784. \end{aligned}

To four decimal places, these probabilities are 0.1296 and 0.7840. The method avoids counting separate cases for success on attempts 1, 2, and 3 when the complement is simpler.

Common mistake avoided: “At least five attempts are needed” includes a first success on attempt 5. The incorrect calculation $P(T>5)=(0.60)^5=0.07776$ instead requires the first five attempts to fail. The correct event is $T\geq 5$, giving $0.60^4=0.1296$.

Worked Example: Repeated Tests After Negative Results

A hypothetical screening test is repeated under identical conditions. Each test independently gives the target positive result with probability 0.25. Let $T$ be the test number of the first positive result. Given that the first three tests were negative, find (a) the probability that the first positive is on one of tests 4, 5, or 6, and (b) the probability that it is exactly on test 5.

Model and method. A positive result is success; $p=0.25$ and $q=0.75$. Thus $T$ is geometric. There is no interest or mortality assumption in this test example. For each requested probability, condition on the stated event $T>3$, which is precisely the information that the first three tests were negative.

(a) First positive on tests 4 through 6. The desired conditional probability is the joint event $3<T\leq 6$, divided by the probability of the conditioning event:

\begin{aligned} P(3<T\leq 6\mid T>3) &=\frac{P(3<T\leq 6)}{P(T>3)}\\ &=\frac{q^3-q^6}{q^3}\\ &=1-q^3\\ &=1-(0.75)^3\\ &=1-0.421875\\ &=0.578125. \end{aligned}

Rounded to four decimal places, this is 0.5781.

(b) First positive exactly on test 5. Given three initial negatives, test 4 must be negative and test 5 positive. Equivalently, divide the joint probability $P(T=5)$ by $P(T>3)$:

\begin{aligned} P(T=5\mid T>3) &=\frac{q^4p}{q^3}\\ &=qp\\ &=(0.75)(0.25)\\ &=0.1875. \end{aligned}

Common mistake avoided: The condition says the first three tests have already been negative; it does not ask for the unconditional probability of a first positive between tests 4 and 6. The conditional calculation divides by $P(T>3)$.

A Reliable Translation Procedure

When the wording is dense, write the event before substituting any numbers. The following sequence is useful for inspections, attempts, repeated tests, and similar first-success settings.

Define success.
State the particular outcome being tracked, such as a defect being found, a task being completed, or a test being positive.

Define the trial count.
Let $T$ be the trial number of the first success, and make clear that the first trial is numbered 1.

Translate boundary words.
Write an event such as $T= k$, $T\leq n$, $T\geq n$, or $a<T\leq b$. Pay particular attention to whether an endpoint is included.

Choose a formula and calculate.
Use the geometric mass for an exact trial, a tail for no success by a deadline, or a tail difference for an interval. For a conditional event, include the conditioning event in the denominator.

Check the interpretation.
Confirm that the power of $q$ counts the required failures and that a “by” deadline includes success on that deadline.

These steps matter because the formula is only as accurate as the event translation. For example, “no success in the first six trials” is $T>6$, while “at least six trials are needed” is $T\geq 6$. Their probabilities are $q^6$ and $q^5$, respectively. The second event permits success on trial 6; the first does not.

Common Wording Traps

Confusing trials with failures. If success first occurs on trial $k$, there are $k-1$ failures, not $k$.
Reading “at least” as “more than.” At least $n$ trials are needed means $T\geq n$; more than $n$ trials are needed means $T>n$.
Missing an inclusive deadline. “By trial $n$” normally includes trial $n$, so it translates to $T\leq n$.
Ignoring a condition. If the problem states that initial trials failed, calculate a conditional probability when asked what happens next; do not report the unconditional probability for the full sequence.
Applying the geometric formula when its model is absent. A geometric calculation requires independent trials with a constant success probability. If the scenario specifies changing success probabilities or dependent outcomes, the geometric formulas do not directly apply.

Key takeaway: Define the success event, let $T$ count the trial of its first occurrence, and translate the wording into an inequality before choosing a formula. In first-success problems, careful handling of $<$, $\leq$, $>$, and $\geq$ is as important as the arithmetic.

Check Your Understanding

For each problem, identify the first-success event before calculating. Assume independent trials with the stated constant success probability.

A hypothetical scanner flags a target feature with probability 0.15 on each scan. What is the probability that it first flags the feature on scan 4? (Answer: 0.0921)
A repeated setup procedure succeeds with probability 0.30 per attempt. What is the probability that it succeeds by attempt 5? (Answer: 0.8319)
An inspection has probability 0.20 of finding the specified issue on each item. What is the probability that at least 6 inspections are required to find one? (Answer: 0.3277)
A diagnostic test returns the target result with probability 0.40 per independent test. What is the probability that the first target result occurs from test 3 through test 6, inclusive? (Answer: 0.3133)
A repeated test has success probability 0.25, and the first four tests are known to have failed. What is the conditional probability that the first success occurs within the next three tests? (Answer: 0.5781)

Wording or event	Event in terms of \(T\)	Probability
First success on trial \(k\)	\(T=k\)	\(q^{k-1}p\)
Success by trial \(n\), inclusive	\(T\leq n\)	\(1-q^n\)
No success in the first \(n\) trials	\(T>n\)	\(q^n\)
At least \(n\) trials are needed	\(T\geq n\)	\(q^{n-1}\)
First success from trial \(a\) through trial \(b\), inclusive	\(a\leq T\leq b\)	\(q^{a-1}-q^b\)
First success after trial \(a\), but by trial \(b\)	\(a<T\leq b\)	\(q^a-q^b\)

What You'll Learn