What's the difference between Geometric Probability and Binomial Probability?

Geometric Probability finds the probability of the first success on a specific trial number; Binomial Probability finds the probability of getting a specific count of successes in a fixed number of trials. Geometric has no upper limit on trials, while Binomial does.

Does X = 0 work in the Geometric Distribution?

No. The first success cannot occur on trial 0 (there is no trial 0). X must be a positive integer: 1, 2, 3, and so on. Some textbooks define a variant that counts failures before the first success, which allows X = 0; always check your textbook's definition.

Why is there a (1-p)^{k-1} in the formula?

The first k-1 trials must fail (each with probability 1-p), and then trial k must succeed (probability p). Multiplying these independent events gives (1-p)^{k-1} \times p.

What if my success probability is very small, like p = 0.01?

The expected value becomes 1/p = 1/0.01 = 100, meaning you typically wait about 100 trials for the first success. The formula still works, but the probability mass spreads over many more trials, so any single P(X=k) will be quite small.

Geometric Probability

Geometric Probability tells you the probability that the first success in a series of independent trials occurs on exactly the k-th attempt.

P(X = k) = (1-p)^{\,k-1}p

Solve a problem with Geometric Probability

Type the problem. The solver will use Geometric Probability where Geometric Probability is the right tool, and tell you when it is not.

What each symbol means

What Geometric Probability takes

Symbol	Meaning
$X$	A random variable representing the trial number on which the first success occurs; X must be a positive integer (1, 2, 3, ...). If you interpret X as counting something other than the trial number of first success—like the number of failures—the formula changes.
$k$	The specific trial number you're asking about (a positive integer: k = 1, 2, 3, ...). If k is zero, negative, or fractional, the formula does not apply to trial-based experiments.
$p$	The probability of success on any single trial, strictly between 0 and 1 (0 < p < 1). If you confuse p with the failure probability $1-p$ , you will invert your answer.

When to use it

Use this when you're repeating the same trial over and over and want to know the probability that your first success lands on a specific trial number.

Level

Usually taught in: Statistics

Worked examples

1. Rolling a die until the first six

Problem

What is the probability that the first six appears on the third roll of a fair die?

$p = \frac{1}{6}, \quad k = 3$
The probability of rolling a 6 on any roll is $1/6$ , and we want to know when the first 6 occurs (on trial 3).
$P(X = 3) = (1 - p)^{k-1} \cdot p$
We apply the Geometric Distribution formula directly.
$P(X = 3) = \left(1 - \frac{1}{6}\right)^{3-1} \cdot \frac{1}{6}$
Substitute $p = 1/6$ and $k = 3$ .
$P(X = 3) = \left(\frac{5}{6}\right)^{2} \cdot \frac{1}{6}$
Simplify: $1 - 1/6 = 5/6$ and $3 - 1 = 2$ .
$P(X = 3) = \frac{25}{36} \cdot \frac{1}{6}$
Compute $(5/6)^2 = 25/36$ .
$P(X = 3) = \frac{25}{216}$
Multiply: $25/36 \times 1/6 = 25/216 \approx 0.116$ or about 11.6%.

Answer: $P(X = 3) = \frac{25}{216}$

This is the basic drill: you need the first two rolls to NOT be a 6 (each with probability $5/6$ ) and then the third roll to BE a 6 (probability $1/6$ ). Since these events are independent, you multiply them to get $(5/6)^2 \times (1/6)$ , which is exactly what the formula gives.

2. Guessing on a multiple-choice test

Problem

A student guesses randomly on a multiple-choice test where each question has four choices. What is the probability that the first correct answer appears on the fifth question?

$p = \frac{1}{4}, \quad k = 5$
With four choices, the guessing probability is $1/4 = 0.25$ . We want the first correct guess on trial 5.
$P(X = 5) = \left(1 - \frac{1}{4}\right)^{5-1} \cdot \frac{1}{4}$
Apply the Geometric formula with these values.
$\left(\frac{3}{4}\right)^{4} \cdot \frac{1}{4}$
Simplify: $1 - 1/4 = 3/4$ and $5 - 1 = 4$ .
$\frac{81}{256} \cdot \frac{1}{4}$
Compute $(3/4)^4 = (3^4)/(4^4) = 81/256$ .
$\frac{81}{1024}$
Multiply: $81/256 \times 1/4 = 81/1024 \approx 0.0791$ .

Answer: $P(X = 5) = \frac{81}{1024} \approx 0.0791$

This harder example requires computing $(3/4)^4$ and handling the larger exponent with care. The probability is lower (about 7.9%) than the die example because we're asking about a later trial (5th instead of 3rd) and the success rate is lower (1/4 instead of 1/6). The student must guess wrong four times, each time with probability $3/4$ , before finally guessing right.

3. Customer support connection attempts

Problem

Your phone plan costs

50/month with

2 per extra gigabyte. The customer support line connects successfully on 20% of calls. What is the probability that your first successful connection occurs on your fourth attempt?

$p = \frac{1}{5}, \quad k = 4$
The connection success rate is 20%, which equals $1/5$ . We want the first success on trial 4.
$P(X = 4) = \left(1 - \frac{1}{5}\right)^{4-1} \cdot \frac{1}{5}$
Substitute into the Geometric formula.
$\left(\frac{4}{5}\right)^{3} \cdot \frac{1}{5}$
Simplify: $1 - 1/5 = 4/5$ and $4 - 1 = 3$ .
$\frac{64}{125} \cdot \frac{1}{5}$
Compute $(4/5)^3 = 64/125$ since $4^3 = 64$ and $5^3 = 125$ .
$\frac{64}{625}$
Multiply: $64/125 \times 1/5 = 64/625 = 0.1024$ .

Answer: $P(X = 4) = \frac{64}{625} = 0.1024$

This real-world scenario involves repeated independent attempts where each has the same success probability. Three calls must fail (each with probability $4/5$ ) before the fourth call succeeds (probability $1/5$ ). The Geometric Distribution applies perfectly because the calls are independent and the 20% success rate is constant across all attempts.

Common mistakes

Where Geometric Probability usually goes wrong

The mistake	Why it is wrong	The fix
I need k successes in a row, then one failure at the end	This confuses Geometric Probability with other scenarios; the formula counts trials until the FIRST success, not multiple successes followed by a failure.	The formula $(1-p)^{k-1}p$ means k-1 consecutive failures followed by exactly 1 success on trial k.
The answer is $p^k$	This ignores the factor $(1-p)^{k-1}$ for the failures and massively overestimates the probability of seeing the first success on a later trial.	Always include both parts: $(1-p)^{k-1}$ for the k-1 failures and $p$ for the success on trial k.
I can use this formula even if the success probability changes with each trial	The Geometric Distribution requires constant probability p; if p changes (e.g., drawing without replacement, or difficulty changing), the formula breaks down.	Only apply this formula when each trial is independent and has exactly the same probability p of success.

Tips and when to use something else

The sum of all probabilities $P(X=1) + P(X=2) + P(X=3) + \ldots$ equals 1; this is why the formula must have the $(1-p)^{k-1}$ factor—it ensures the distribution is valid.
A larger p (higher success rate) means probabilities concentrate near k=1; a smaller p spreads the distribution over many trials. With p very close to 0, first success might take hundreds or thousands of trials.
If you're counting the number of successes in a fixed number of trials (not trials until first success), use Binomial Probability instead.
The expected value (mean) of a geometric distribution is $1/p$ ; with $p = 0.2$ , you expect to wait an average of 5 trials for the first success.

Related formulas

Binomial ProbabilityBinomial Probability finds the likelihood of exactly k successes in n independent identical trials, each with fixed probability p.Basic ProbabilityBasic Probability calculates the likelihood of an event by dividing favorable outcomes by the total number of possible outcomes.Multiplication RuleFind the probability that two events both occur when one might affect the chance of the other using multiplication of probabilities.Independent EventsWhen events are independent, multiply their individual probabilities together to find the probability that both events occur.Expected ValueExpected value is the average outcome you'd get if you repeated an experiment many times, weighted by how likely each outcome is.

Frequently asked questions

What's the difference between Geometric Probability and Binomial Probability?: Geometric Probability finds the probability of the first success on a specific trial number; Binomial Probability finds the probability of getting a specific count of successes in a fixed number of trials. Geometric has no upper limit on trials, while Binomial does.
Does X = 0 work in the Geometric Distribution?: No. The first success cannot occur on trial 0 (there is no trial 0). X must be a positive integer: 1, 2, 3, and so on. Some textbooks define a variant that counts failures before the first success, which allows X = 0; always check your textbook's definition.
Why is there a $(1-p)^{k-1}$ in the formula?: The first k-1 trials must fail (each with probability $1-p$ ), and then trial k must succeed (probability $p$ ). Multiplying these independent events gives $(1-p)^{k-1} \times p$ .
What if my success probability is very small, like p = 0.01?: The expected value becomes $1/p = 1/0.01 = 100$ , meaning you typically wait about 100 trials for the first success. The formula still works, but the probability mass spreads over many more trials, so any single $P(X=k)$ will be quite small.

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Reviewed 2026-09-18