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The Geometric Probability Mass Function

Learn to identify and calculate the probability that the first success occurs on a specified trial.

Intermediate 9 min read

What You'll Learn

  • State the assumptions for a geometric waiting-time model.
  • Define the random variable for the trial of the first success.
  • Derive the probability mass function from the required trial sequence.
  • Distinguish the k−1 failures before the final success on trial k.
  • Calculate exact first-success probabilities and interpret them in context.
  • Check that the formula assigns probabilities to the possible waiting times.

From a Waiting Time to an Exact Probability

A geometric waiting-time question asks how long it takes for a success to occur for the first time. For example, a hypothetical device might pass a check with the same probability on each independent attempt. If \(T\) records the attempt on which it first passes, asking for \(P(T=4)\) means asking for the probability that the first three attempts fail and the fourth succeeds.

That event has a specific order: failures first, then one success. Keeping track of this order is the key to the geometric probability mass function. In particular, the first \(k-1\) trials must fail for the first success to occur on trial \(k\); trial \(k\) itself must succeed.

The Geometric Model and Its Probability Mass Function

Use a geometric model when a sequence of trials meets the Bernoulli-trial conditions: each trial has two outcomes, success or failure; the trials are independent; and the probability of success is the same on every trial. Let \(p\) be that success probability, with \(0<p<1\). Continue the trials until the first success.

Definition: The random variable \(T\) is the trial number on which the first success occurs. It can take the positive integer values \(1,2,3,\ldots\). If each independent trial has success probability \(p\), then the geometric probability mass function is $$ P(T=k)=(1-p)^{k-1}p,\qquad k=1,2,3,\ldots $$ Here \(1-p\) is the probability of failure on any one trial.

A probability mass function gives the probability assigned to each possible value of a discrete random variable. In this model, \(P(T=k)\) is the probability that the first success occurs exactly on trial \(k\), not merely that trial \(k\) is a success. The distinction matters: trial \(k\) could succeed even if an earlier trial had already succeeded, which would make \(T\) smaller than \(k\).

Why the Exponent Is \(k-1\)

The event \(T=k\) requires a failure on each of trials \(1\) through \(k-1\), followed by a success on trial \(k\). There are \(k-1\) required failures and one required success. By independence, the probability of this particular sequence is the product of the probabilities of its outcomes.

$$ \begin{aligned} P(T=k) &=\underbrace{(1-p)(1-p)\cdots(1-p)}_{k-1\text{ failures}}\;p\\ &=(1-p)^{k-1}p. \end{aligned} $$

This also shows why the exponent is not \(k\): only \(k-1\) failures are required before the first success. The final factor \(p\) is not part of the failure run; it accounts for the success on trial \(k\).

For \(k=1\), there are no preceding failures. The event \(T=1\) is simply a success on the first trial, so \(P(T=1)=p\). This agrees with the formula because \((1-p)^0p=p\). For larger \(k\), each additional wait requires one more failure before the success.

Worked Example: First Pass on the Fourth Check

Suppose, hypothetically, that a device passes each independent check with probability \(p=0.20\). Let \(T\) be the number of the first check it passes. Find the probability that it first passes check 4.

Model check. The situation specifies two outcomes on each check (pass or fail), independent checks, and the same pass probability on every check. These are the geometric-model conditions. The requested event requires three failures, then a pass.

Using the geometric probability mass function with \(k=4\) and \(p=0.20\):

$$ \begin{aligned} P(T=4) &=(1-p)^{4-1}p\\ &=(1-0.20)^3(0.20)\\ &=(0.80)^3(0.20)\\ &=0.512(0.20)\\ &=0.1024. \end{aligned} $$

The probability is \(0.1024\), or \(10.24\%\). In context, about \(10.24\%\) of sequences of checks under this model have the first pass on check 4. The three failures are checks 1, 2, and 3; the pass is check 4.

Reading the Formula as a Trial Sequence

The formula is easiest to use reliably when each factor is matched to a part of the event. First identify what counts as success, and then state the event \(T=k\) in words. Count the failures that must occur before the first success, use \(1-p\) for each one, and use \(p\) for the success on trial \(k\). Independence permits multiplying these probabilities.

1
Identify the success and \(p\).
Use the success definition for the setting and the probability of success on one trial.
2
Translate the requested wait.
For \(T=k\), trials 1 through \(k-1\) fail and trial \(k\) succeeds.
3
Multiply the outcome probabilities.
There are \(k-1\) factors of \(1-p\) and one final factor of \(p\), giving \((1-p)^{k-1}p\).

Worked Example: A Positive Result on the Second Test

In a hypothetical testing procedure, each independent test gives a positive result with probability \(p=0.60\). Let \(T\) be the number of the first test that is positive. Find \(P(T=2)\).

Model check. Each test has two outcomes for this question (positive or not positive), and the scenario assumes independent tests with a constant positive-result probability of \(0.60\). The event \(T=2\) is a negative result on test 1 followed by a positive result on test 2.

$$ \begin{aligned} P(T=2) &=(1-p)^{2-1}p\\ &=(1-0.60)^1(0.60)\\ &=(0.40)(0.60)\\ &=0.24. \end{aligned} $$

Thus, the probability that the first positive result occurs on test 2 is \(0.24\), or \(24\%\). The exponent is 1 because exactly one earlier test must be negative; the second test must be positive.

Checking That the Probabilities Form a Distribution

A probability mass function for \(T\) must assign nonnegative probabilities to its possible values, and those probabilities must sum to 1. For \(0<p<1\), each value \((1-p)^{k-1}p\) is nonnegative. To check the total, set \(r=1-p\), so \(0<r<1\), and use the infinite geometric-series identity \(\sum_{j=0}^{\infty}r^j=1/(1-r)\):

$$ \begin{aligned} \sum_{k=1}^{\infty}P(T=k) &=\sum_{k=1}^{\infty}(1-p)^{k-1}p\\ &=p\sum_{j=0}^{\infty}(1-p)^j\\ &=p\left(\frac{1}{1-(1-p)}\right)\\ &=p\left(\frac{1}{p}\right)\\ &=1. \end{aligned} $$

The index change in the second line uses \(j=k-1\), so that \(j\) begins at 0. This calculation confirms that, under the model with \(0<p<1\), the probabilities across all finite first-success trial numbers sum to 1.

If \(p=1\), success occurs on the first trial with probability 1, so \(P(T=1)=1\) and \(P(T=k)=0\) for \(k>1\). If \(p=0\), a success never occurs, so there is no finite first-success trial; the usual geometric waiting-time model excludes that case.

Worked Example: First Pass on the Fifth Inspection

Suppose, hypothetically, that a part passes an inspection with probability \(p=0.15\) on each independent inspection. Let \(T\) be the inspection number on which it first passes. Calculate the probability that it first passes on inspection 5.

Model check. Inspection outcomes are pass or fail, and the scenario assumes independent inspections with a constant pass probability of \(0.15\). For \(T=5\), inspections 1 through 4 must fail and inspection 5 must pass.

$$ \begin{aligned} P(T=5) &=(1-p)^{5-1}p\\ &=(1-0.15)^4(0.15)\\ &=(0.85)^4(0.15)\\ &=0.52200625(0.15)\\ &=0.0783009375. \end{aligned} $$

Rounded to four significant figures, the probability is \(0.07830\), or about \(7.830\%\). This is the probability of four failures followed by a pass, not simply the probability of passing inspection 5.

Common Mistakes and Why the Model Matters

  • Using \(k\) failures. The mistaken reasoning is that a wait of \(k\) trials means multiplying by \(1-p\) a total of \(k\) times. But if all \(k\) trials fail, there is no success on trial \(k\). The event \(T=k\) has \(k-1\) failures and one success, so its probability is \((1-p)^{k-1}p\).
  • Leaving out the final \(p\). The probability \((1-p)^{k-1}\) only describes failures on the first \(k-1\) trials. It does not require success on trial \(k\). Multiply by \(p\) to specify that final success.
  • Counting any success on trial \(k\) as \(T=k\). A success on trial \(k\) is not necessarily the first success. For example, success on both trials 1 and 2 means \(T=1\), not \(T=2\). The preceding failures are what make the success on trial \(k\) the first one.
  • Using the formula when the assumptions fail. If success probabilities change from trial to trial or outcomes are dependent, multiplying the same failure probability \(k-1\) times is not justified. Check the trial conditions before using the geometric probability mass function.

The geometric probability mass function is useful whenever the question concerns an exact first-success trial under a stable sequence of independent trials. It converts the event into a simple ordered sequence, making the calculation transparent: the waiting portion consists of failures, and the final trial supplies the success.

Check Your Understanding

For each question, identify the required sequence of outcomes before calculating or explaining.

  1. If \(p=0.30\), write the event described by \(T=3\) in terms of failures and success, then calculate \(P(T=3)\).
  2. A learner says, “The probability that the first success is on trial 6 is \((1-p)^6\).” Identify the error and give the correct expression.
  3. In a hypothetical setting with \(p=0.45\), find the probability that the first success occurs on trial 1. Explain why the exponent is zero.
  4. Why is a constant success probability needed to use the formula \((1-p)^{k-1}p\)?
  5. For \(0<p<1\), what does the sum \(\sum_{k=1}^{\infty}P(T=k)=1\) verify about the geometric probability mass function?

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