Geometric Problems With First-Success Waiting Times

Turning a Waiting-Time Question Into a Probability

A question about repeated attempts can ask for quite different things: whether the first success happens on trial 5, whether it happens within 5 trials, or whether it takes longer than 5 trials. These events overlap in everyday language, but they are not the same probability. The geometric model gives a direct way to calculate each one once the trial process and the success probability are specified.

In the geometric model, $X$ is the trial number on which the first success occurs. The model requires a sequence of trials with two outcomes, a constant probability $p$ of success on every trial, and independence between trials. Thus the probability of failure on each trial is $1-p$, and previous failures do not alter the chance of success on the next trial.

Definition: For independent trials with constant success probability $p$, where $0<p\leq 1$, the geometric random variable $X$ is the trial number of the first success. For an integer $k\geq 1$, $$ P(X=k)=(1-p)^{k-1}p. $$ The first $k-1$ trials must be failures, and trial $k$ must be a success.

The repeated-trial assumptions matter. A numerical calculation can be carried out from a stated $p$, but its interpretation as a geometric probability is justified only if the trials are independent and the success probability remains constant. In the examples below, those conditions are explicitly part of the hypothetical setup.

First Success on a Specified Trial

To find the probability that the first success is on trial $k$, count the required pattern: $k-1$ failures followed by one success. Independence lets us multiply the trial probabilities. The order is essential; success on an earlier trial would mean that the first success was not on trial $k$.

Worked Example: First Successful Check on Trial 5

Suppose a hypothetical device is checked repeatedly. Each check independently detects a fault with probability $p=0.25$, with the same probability on every check. Let $X$ be the number of checks through and including the first detection. Find the probability that the first detection occurs on check 5.

The model assumptions are stipulated: each check has two outcomes (detection or no detection), checks are independent, and the detection probability is constant at 0.25. A first detection on check 5 requires four checks without detection, followed by a detection.

P(X=5)=(1-p)^{5-1}p=(1-0.25)^4(0.25)=0.75^4(0.25)

Since $0.75^4=0.31640625$,

P(X=5)=0.31640625(0.25)=0.0791015625\approx 0.0791.

Thus, under this model, there is about a 7.91% chance that the first fault detection occurs on the fifth check. This is an exact-trial probability, not the probability of at least one detection during the first five checks.

The exponent is one less than the trial number because the success trial itself is not a failure. For trial 1, for example, there are zero preceding failures, so $P(X=1)=p$.

Success Within a Deadline

A deadline of $m$ trials includes trials 1 through $m$. The event “success within $m$ trials” is $X\leq m$. A simple way to calculate it is to use its complement: the first success does not occur within the deadline exactly when all of the first $m$ trials are failures. Independence makes the probability of those $m$ failures $(1-p)^m$.

Result: For a positive integer deadline $m$, $$ P(X\leq m)=1-(1-p)^m. $$ The complementary probability is $$ P(X>m)=(1-p)^m. $$ Here $X>m$ means the first success occurs after all of the first $m$ trials have failed.

The cumulative formula can also be derived by adding the mutually exclusive exact-trial probabilities. Writing $q=1-p$, the first-success events $X=1,\ldots,X=m$ partition the event $X\leq m$. Therefore,

\begin{aligned} P(X\leq m) &=\sum_{k=1}^{m}P(X=k)\\ &=\sum_{k=1}^{m}q^{k-1}p\\ &=p\frac{1-q^m}{1-q}\\ &=1-q^m\\ &=1-(1-p)^m. \end{aligned}

The third line uses the finite geometric-sum formula, and the fourth uses $1-q=p$. This derivation also confirms why the deadline formula includes every possible first-success trial from 1 through $m$, rather than only the last trial.

Worked Example: Success by a Six-Trial Deadline

In a hypothetical quality-control process, each independently tested item has probability $p=0.30$ of meeting a specified criterion. Assume this probability is constant across tests. Let $X$ be the number of tests until the first item meets the criterion. Find the chance of success within the first six tests.

The stated process satisfies the geometric assumptions: each test is treated as a success or failure, outcomes are independent, and the probability of success is 0.30 on every test. “Within the first six” means $X\leq 6$. It is easier to subtract the probability that all six tests fail:

\begin{aligned} P(X\leq 6) &=1-(1-p)^6\\ &=1-(1-0.30)^6\\ &=1-0.70^6\\ &=1-0.117649\\ &=0.882351\approx 0.8824. \end{aligned}

The model assigns about an 88.24% chance that the first item meeting the criterion appears on one of tests 1 through 6. The remaining 11.76% is the probability that all six tested items fail to meet it, so the first success is later than test 6.

Waiting Beyond a Deadline

“After $m$ trials” means that the first success has not happened on any of trials 1 through $m$, or $X>m$. This event requires $m$ consecutive failures. The deadline trial itself is included among those failures; success on trial $m$ is still success within the deadline, not beyond it.

Worked Example: First Success After an Eight-Trial Deadline

Suppose a hypothetical sensor has probability $p=0.12$ of registering a signal on each activation. Activations are independent, and the signal probability stays at 0.12 each time. Let $X$ be the activation on which the first signal registers. Find the probability that the first signal comes after activation 8.

The model is geometric under the given assumptions. To have $X>8$, the sensor must fail to register a signal on each of the first eight activations. Each failure has probability $1-0.12=0.88$, so

\begin{aligned} P(X>8) &=(1-p)^8\\ &=(1-0.12)^8\\ &=0.88^8\\ &\approx 0.3596. \end{aligned}

Thus there is about a 35.96% chance that the first registered signal occurs after the first eight activations. Equivalently, the chance of at least one signal by activation 8 is $1-0.3596345\approx 0.6404$, or about 64.04%.

The two results are complements: $P(X\leq 8)+P(X>8)=1$. The “after 8” event does not mean that a signal must occur on activation 9; the first signal could occur on activation 9 or on any later activation.

A Reliable Calculation Process

Before substituting into a formula, translate the wording into an event involving $X$. Then select the formula that matches that event. The following sequence helps keep the trial number and deadline straight.

Define the waiting time.
State that $X$ is the trial number of the first success, and identify what counts as a success in context.

Check the model conditions.
Confirm that trials are independent, have two relevant outcomes, and share a constant success probability $p$. If these conditions are not appropriate, the geometric calculation is not justified.

Translate the question.
“On trial $k$” is $X=k$; “by trial $m$” is $X\leq m$; “after trial $m$” is $X>m$.

Calculate and interpret.
Use the matching formula, show the values substituted, and describe the resulting probability in the original units and context.

A useful check is to make sure the probability matches the event’s scope. An exact-trial probability requires one particular pattern of failures followed by success. A by-deadline probability includes multiple possible first-success trials. A beyond-deadline probability requires failures throughout the deadline.

Why the Wording Matters

A common mistake is to see a deadline of $m$ trials and calculate $P(X=m)$ when the question asks for success within $m$. That answers a narrower question: first success on the final trial only. For example, $P(X=6)$ requires failures on trials 1 through 5 and success on trial 6; it excludes first success on trials 1 through 5. By contrast, $P(X\leq 6)$ includes all six possibilities.

The reverse mistake is also possible: using $P(X\leq m)$ when asked whether the first success occurs after $m$. The corrected event is $X>m$, with probability $(1-p)^m$, not $1-(1-p)^m$. Checking whether the wording says “by” or “after” prevents this complement error.

Key takeaway: For a first-success waiting time, “on trial $k$” means $P(X=k)=(1-p)^{k-1}p$; “within $m$ trials” means $P(X\leq m)=1-(1-p)^m$; and “beyond $m$ trials” means $P(X>m)=(1-p)^m$. Interpret the event before calculating.

Check Your Understanding

For each question, identify the event first, then choose a geometric probability expression.

If $p=0.20$, what expression gives the probability that the first success occurs on trial 4?
If $p=0.20$, what expression gives the probability of at least one success within the first 4 trials?
Explain why the probability of a first success after trial 4 is $(1-p)^4$, rather than $(1-p)^3$.
A hypothetical process has independent trials and constant success probability $p=0.10$. What is the probability that the first success occurs by trial 3? Give a contextual interpretation if each trial is one machine activation.
State the independence and constant-probability assumptions needed before treating a waiting time as geometric.