Geometric Tail Probabilities

Why Geometric Tails Are Useful

A geometric model describes the number of trials needed to obtain the first success when each trial has the same success probability and trials are independent. Often, the question is not about one exact waiting time. Instead, it asks whether success will take longer than a specified number of trials, or whether it will happen within that many trials. These are tail and cumulative probabilities.

For example, a quality-control team might ask how likely it is that a randomly selected acceptable part will not be found in the first several inspections. A probability of waiting beyond five trials is different from the probability of first succeeding on trial five. Keeping those events separate is essential for a correct calculation.

Definition: Let \(X\) be the trial number on which the first success occurs in independent Bernoulli trials, each with success probability \(p\), where \(0<p<1\). Then \(X\) has a geometric distribution with probability mass function \(P(X=x)=(1-p)^{x-1}p\) for \(x=1,2,3,\ldots\). For a nonnegative integer \(k\), the event \(X>k\) means that the first \(k\) trials are all failures; \(X\leq k\) means that a success occurs on or before trial \(k\).

Waiting Beyond the First \(k\) Trials

To wait beyond \(k\) trials, the first success must not occur on any of the first \(k\) trials. In other words, all \(k\) of those trials must be failures. A single failure has probability \(1-p\), and independence means the probability of failures on all \(k\) trials is the product of their probabilities.

$$ P(X>k)=(1-p)^k $$

This expression applies for an integer \(k\geq 0\). At \(k=0\), no trials have been performed, so \(P(X>0)=1\), consistent with \((1-p)^0=1\). When \(k\) increases, the event becomes more restrictive: to wait beyond more trials, more consecutive failures are required.

The phrase “beyond \(k\) trials” has a precise boundary: the first \(k\) trials are failures, and the first success, if it occurs, is on trial \(k+1\) or later. This is not the same as “the first success is on trial \(k\) or later.” That second event is \(X\geq k\), which requires only the first \(k-1\) trials to be failures:

P(X\geq k)=(1-p)^{k-1}, \qquad k=1,2,3,\ldots

Worked Example: Waiting Beyond Five Inspections

Suppose, hypothetically, that an inspection has a probability \(p=0.20\) of finding the target defect on any one item examined. Let \(X\) be the inspection number on which the first target defect is found. Find the probability that the first five inspections do not find it.

Check the model. The geometric calculation assumes that each inspection has the same probability \(0.20\) of finding the defect and that the inspection outcomes are independent. We use those assumptions for this hypothetical process.

The event that the first five inspections do not find the defect is \(X>5\). The probability of failure on one inspection is \(1-p=1-0.20=0.80\), so:

$$ P(X>5)=(1-p)^5=(0.80)^5=0.32768 $$

Thus, the model assigns probability \(0.3277\), or about \(32.77\%\), to not finding the target defect in the first five inspections. This is a probability of a waiting-time event, not the probability that the first defect is found exactly on inspection five.

Succeeding Within \(k\) Trials

The event \(X\leq k\) means that at least one success occurs among the first \(k\) trials. Its complement is \(X>k\), meaning all of those trials are failures. Since an event and its complement have probabilities that add to one, the probability of success within \(k\) trials is:

P(X\leq k)=1-P(X>k)=1-(1-p)^k

This complement calculation is often simpler than adding the probabilities of success on trial 1, trial 2, and so on. It also gives a useful check: \(P(X\leq k)\) and \(P(X>k)\) must add to 1. The formula includes success on trial \(k\), because “within \(k\)” means on any of trials 1 through \(k\), inclusive.

Worked Example: Success Within Four Attempts

Imagine a hypothetical device that completes a task successfully on each attempt with probability \(p=0.35\). Assume attempts are independent and the success probability remains constant. What is the probability of at least one successful completion within four attempts?

Let \(X\) be the attempt on which the first successful completion occurs. “Within four attempts” is \(X\leq 4\). It is easier to calculate the complementary event: failure on all four attempts.

P(X\leq 4)=1-P(X>4)=1-(1-0.35)^4

Substituting the failure probability and carrying out the arithmetic gives:

P(X\leq 4)=1-(0.65)^4 =1-0.17850625 =0.82149375

The probability of at least one successful completion in the first four attempts is approximately \(0.8215\), or \(82.15\%\). The complementary probability of four failures is approximately \(0.1785\), so the two probabilities sum to 1, as required.

Justifying the Cumulative Formula

The complement argument establishes the formula directly from the trial process. It can also be checked by adding the probabilities of every possible first-success trial from 1 through \(k\). The geometric probability mass function gives \(P(X=j)=(1-p)^{j-1}p\), so:

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

The third line uses the finite geometric-series identity \(1+r+\cdots+r^{k-1}=(1-r^k)/(1-r)\), with \(r=1-p\). Since \(1-(1-p)=p\), the factor \(p\) cancels, giving the same result as the complement method. This calculation also shows that the cumulative probability includes every possible first-success trial from 1 through \(k\).

Worked Example: Distinguishing “At Least” From “Beyond”

Suppose, hypothetically, that each test has an independent probability \(p=0.10\) of detecting a signal, with the same probability on every test. Let \(X\) be the test number of the first detection. Compare the probabilities that the first detection occurs on test 8 or later and that it occurs after the first 8 tests.

On test 8 or later means \(X\geq 8\). Only the first seven tests must be failures, so:

P(X\geq 8)=(1-0.10)^7=(0.90)^7=0.4782969

After the first 8 tests means \(X>8\). The first eight tests must all be failures, so:

$$ P(X>8)=(1-0.10)^8=(0.90)^8=0.43046721 $$

The first probability includes detection on test 8; the second does not. Their difference is exactly the probability of first detection on test 8:

$$ P(X=8)=(0.90)^7(0.10)=0.04782969 $$

Indeed, \(P(X\geq 8)-P(X>8)=0.4782969-0.43046721=0.04782969\). The numbers are consistent because “test 8 or later” includes the single boundary case that “after the first 8 tests” excludes.

Using the Formulas Carefully

Before applying a geometric tail formula, define what counts as a success and what \(X\) records. Then identify the trial-count boundary in the question. The following translations are useful:

Question wording	Event	Probability
No success in the first \(k\) trials; success takes longer than \(k\) trials	\(X>k\)	\((1-p)^k\)
At least one success within the first \(k\) trials	\(X\leq k\)	\(1-(1-p)^k\)
First success occurs on trial \(k\) or later	\(X\geq k\)	\((1-p)^{k-1}\)
First success occurs exactly on trial \(k\)	\(X=k\)	\((1-p)^{k-1}p\)

These expressions are related but answer different questions. In particular, “exactly on trial \(k\)” requires \(k-1\) failures followed by a success, whereas “after the first \(k\) trials” requires \(k\) failures. A quick event description in words often prevents an off-by-one error.

Common Mistakes and Why Tails Matter

Using \(k-1\) failures for \(X>k\). That reasoning leaves open the possibility of success on trial \(k\). For \(X>k\), all \(k\) trials must be failures, so the power is \(k\).
Confusing a tail with an exact probability. \((1-p)^k\) is the probability of no success in the first \(k\) trials, not the probability that the first success occurs exactly on trial \(k\). The exact probability includes a success factor \(p\).
Subtracting in the wrong direction. The chance of success within \(k\) trials is the complement of waiting beyond \(k\), so it is \(1-(1-p)^k\), not \((1-p)^k-1\).
Ignoring the assumptions. If success probabilities change from trial to trial or outcomes are dependent, multiplying a single failure probability \(k\) times may not be justified. The geometric model needs independent trials with a constant success probability.

Tail probabilities are useful because they answer practical threshold questions: how likely is it that a process will take more than a specified number of tries, or how likely is it that a target will be reached by a deadline measured in trials? The formulas also make transparent how the threshold and per-trial success probability affect that chance. A larger \(p\) makes the failure probability \(1-p\) smaller, while a larger \(k\) requires more consecutive failures for \(X>k\).

Key takeaway: In a geometric model, waiting beyond \(k\) trials requires \(k\) consecutive failures, so \(P(X>k)=(1-p)^k\). Success within \(k\) trials is its complement: \(P(X\leq k)=1-(1-p)^k\). Translate the wording into an event before choosing the exponent.

Check Your Understanding

For each question, identify the event before choosing a formula.

A trial has success probability \(p=0.25\). What is the probability that the first success occurs after the first three trials?
Using the same \(p\), what is the probability of at least one success within the first three trials?
Explain why the probability that the first success occurs on trial 6 or later uses five failure probabilities rather than six.
For \(p=0.40\), write an expression for the probability that the first success occurs exactly on trial 4.
State the independence and constant-probability assumptions needed to use \(P(X>k)=(1-p)^k\).

What You'll Learn