0.147
0.657
0.343
3.333333
trials
7.777778
trials²
2.788867
trials
2
trials
2.333333
failures
0.3
0.147
0.657
0.343
3.333333
trials
7.777778
trials²
2.788867
trials
2
trials
2.333333
failures
0.3
The Geometric Distribution Calculator computes probabilities and statistics for the geometric distribution, which models the number of Bernoulli trials needed to obtain the first success. It is the discrete counterpart of the exponential distribution and the only discrete distribution possessing the memoryless property.
If each trial independently succeeds with probability $$p$$, the probability that the first success occurs on trial $$k$$ is given by the probability mass function:
$$P(X = k) = (1 - p)^{k-1} \cdot p, \quad k = 1, 2, 3, \ldots$$
This formula counts $$k - 1$$ failures followed by one success. The cumulative distribution function is:
$$F(k) = P(X \leq k) = 1 - (1 - p)^k$$
The expected number of trials until the first success is $$E[X] = \frac{1}{p}$$. The variance is $$\text{Var}(X) = \frac{1 - p}{p^2}$$. The median is $$\left\lceil \frac{-1}{\log_2(1 - p)} \right\rceil$$. As $$p$$ increases, the distribution becomes more concentrated near $$k = 1$$, since success is more likely on early trials.
The geometric distribution is the only discrete distribution with the memoryless property: $$P(X > s + t \mid X > s) = P(X > t)$$. This means that if you have already failed $$s$$ times, the probability distribution of remaining trials is the same as starting fresh. This is analogous to the memoryless property of the exponential distribution.
The geometric distribution is a special case of the negative binomial distribution with $$r = 1$$ (waiting for the first success). It relates to the binomial distribution: if the number of successes in $$n$$ trials is binomial, then the number of trials until the first success is geometric. As $$p \to 0$$ with $$\lambda = 1/p$$, the geometric distribution approximates the exponential distribution.
The geometric distribution models scenarios such as the number of coin flips until the first heads, the number of items inspected before finding the first defective one, the number of job applications before receiving an offer, or the number of network packet transmissions before a successful delivery.
Enter the success probability $$p$$ (between 0 and 1) and the trial number $$k$$ (positive integer). The calculator computes the PMF, CDF, survival function, mean, variance, standard deviation, and median.
The calculator evaluates $$P(X = k) = (1-p)^{k-1} \cdot p$$ using exponentiation. The CDF is computed as $$1 - (1-p)^k$$. The survival function is $$(1-p)^k$$. Statistical measures are calculated from the standard formulas for the geometric distribution.
The PMF value tells you the probability that the first success occurs on exactly trial $$k$$. The CDF gives the probability of achieving at least one success within $$k$$ trials. The survival function gives the probability that all $$k$$ trials fail. A higher $$p$$ shifts the distribution toward smaller $$k$$ values.
Inputs
Results
The probability of the first heads on the 3rd flip is 12.5%. There is an 87.5% chance of getting at least one heads within 3 flips. On average, you need 2 flips.
Inputs
Results
With a 5% defect rate, the probability of finding the first defective item on exactly the 10th inspection is about 3.2%. There is a 40.1% chance of finding a defect within 10 inspections.
The geometric distribution models the number of independent Bernoulli trials required to achieve the first success. It applies to any scenario with repeated independent trials, each having the same probability $$p$$ of success.
Yes. One convention counts the number of trials until the first success ($$k = 1, 2, 3, \ldots$$), while the other counts the number of failures before the first success ($$k = 0, 1, 2, \ldots$$). This calculator uses the first convention, where $$k$$ starts at 1.
If you have already flipped a coin 10 times without heads, the probability of needing $$t$$ more flips is the same as if you were starting from scratch. Past failures do not change the probability of future success—each trial is independent.
The geometric distribution is a special case of the negative binomial with $$r = 1$$. While the geometric counts trials until the 1st success, the negative binomial counts trials until the $$r$$-th success.
The survival function $$P(X > k) = (1-p)^k$$ gives this directly. It is the probability that all $$k$$ trials result in failure, so the first success has not yet occurred.
If $$p = 1$$, success occurs on the first trial with certainty, so $$P(X = 1) = 1$$. If $$p = 0$$, success never occurs and the distribution is undefined, as the expected value $$1/p$$ diverges to infinity.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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