Time until the next event — and the assumption about wear you are making by using it.
Time until the next event
λ and the mean are reciprocals: λ = 0.50000 means a mean of 2.00000. Swapping them is the classic error and the answer still looks plausible.
λ = 0.50000, mean = 2.00000
0.223130
P(event af3.000er t). The median wait is 1.3863 — shorter than the mean, because the distribution is right-skewed.
Probability
0.223130
after t
Mean
2.00000
1 ÷ λ
Median
1.38629
ln2 ÷ λ — well below the mean
Standard deviation
2.00000
equals the mean, always
It has already run this long without an event
Does that change the wait from here?
Given 10.00 elapsed
0.22313016
another 3.00 with no event
Starting fresh
0.22313016
3.00 with no event
Identical. The exponential has no memory of elapsed time — which means a constant hazard rate, and that a component is exactly as likely to fail in its next hour after 10 hours as when new. That is a strong physical claim, and false for anything that wears out.
What this tool shows
Takes a rate or a mean and always shows both, because they are reciprocals and swapping them gives a plausible-looking wrong answer. The panel below shows that elapsed time changes nothing — which is a constant hazard rate, and false for anything that wears out.
P(before t), P(after t), and between two times
Rate and mean, always shown together
The density drawn with the region shaded
The median, which is well below the mean
A live memorylessness demonstration
When a constant hazard rate is the wrong model
Constant hazard, shown Rate vs mean, both Exact tails Weibull named as the fix
Assumes failure rate never changes with age. Most things wear.
Updated 8 September 2026 · Works in any browser, no installation
P(X > t) = e⁻ᵤᵗ. The survival function is the whole distribution in one expression, and its shape is the claim: the chance of surviving another hour is the same whatever has already elapsed.
At a glance
Formula shown
Density λe^(−λt), CDF 1 − e^(−λt), survival e^(−λt). Mean and standard deviation are both 1/λ; the median is ln2/λ ≈ 0.693/λ, well below the mean. The hazard rate — instantaneous failure risk given survival so far — is constantly λ.
Scenario support
Time between arrivals at a queue; time until a component fails, when failures are random rather than wear-driven; the gap between radioactive decays; service times in a simple queueing model.
Educational estimate
Planning support from the values you enter — not professional advice.
A constant hazard rate is a claim about your process
The exponential is the continuous memoryless distribution, and the physical translation of that is sharper than the mathematical statement.
The hazard rate is constant. The instantaneous risk of failing in the next moment, given that you have survived so far, is λ — and it never changes. A component that has run for a thousand hours is exactly as likely to fail in its next hour as one taken out of the box.
State it that way and it becomes obviously false for most physical things. Bearings wear. Batteries degrade. Materials fatigue. For any of those the hazard rises with age, and an exponential model will systematically underestimate late-life failures — which is exactly the period maintenance planning cares about.
The classic picture is the bathtub curve: high hazard early from manufacturing defects, low and roughly flat in the middle, rising at the end from wear. The exponential describes the flat middle and nothing else.
Where it genuinely holds: processes with no memory in the physical sense. Radioactive decay is the pure case — an atom does not age. Arrivals at a queue from a large independent population. Failures caused by external shocks rather than internal wear, where a random event either arrives or does not.
The tool’s panel makes this checkable: set the elapsed time to anything you like and the remaining wait does not move. If that is not true of your process, the exponential is the wrong model however convenient its arithmetic.
The standard replacement is the Weibull distribution, which adds a shape parameter controlling whether the hazard rises, falls or stays flat. Shape = 1 is the exponential, so it is a strict generalisation.
Rate or mean — they are reciprocals
The commonest arithmetic error with this distribution, and it produces answers that look entirely reasonable.
λ is a rate: events per unit time. “Four calls per hour” means λ = 4.
1/λ is a mean: time per event. Four calls an hour means a mean gap of 0.25 hours.
Enter 4 where 0.25 belongs and every probability is wrong — but plausibly wrong, because the output is still a number between 0 and 1. Nothing flags it. The tool takes either and prints both, so the reciprocal is always visible.
The Poisson connection makes this concrete and is worth holding onto: if events arrive as a Poisson process at rate λ per hour, then the number in an hour is Poisson(λ) and the gap between them is Exponential(λ). Same λ, same process, two views.
Units matter for the same reason. A rate of 4 per hour is 0.0667 per minute, and mixing time units in with the same λ is the second version of this error.
The median is well below the mean
The exponential is strongly right-skewed, and the gap between its centres is large enough to change how you should plan.
Mean = 1/λ. Median = ln2/λ ≈ 0.693/λ. The median is about 69% of the mean, always — the ratio does not depend on λ.
So about 63% of waits are shorter than the mean: P(X < 1/λ) = 1 − e⁻¹ = 0.632, whatever the rate. Most waits are below average and a few long ones pull it up.
The same 63% appeared on the geometric page for budgeting the mean number of trials, and it is the same constant for the same reason — the two distributions are the discrete and continuous versions of one idea.
The standard deviation equals the mean, which is unusual and diagnostic. A coefficient of variation of exactly 1 is the exponential’s signature: if your waiting-time data has a CV far from 1, it is not exponential. Above 1 suggests clustering or a mixture; below 1 suggests something more regular than random.
For planning, quote a high percentile rather than the mean. The 95th is ln(20)/λ ≈ 3/λ — three times the mean — which is the same 3/p heuristic the geometric page arrives at.
Why queueing theory is built on it
The exponential dominates queueing models, and the reason is tractability rather than realism — which is worth knowing before trusting the output.
Memorylessness means the future of a queue depends only on its current state, not on how long anyone has been waiting. That makes the system a Markov chain, and Markov chains have closed-form answers. Drop the assumption and most of those formulas disappear.
Arrivals are the safer half. When customers arrive independently from a large population, the gaps really are close to exponential. This is well supported empirically.
Service times are the weaker half. An exponential service time says a call that has already run twenty minutes is as likely to end in the next minute as one that just started. Real service times are usually far more regular than that, and treating them as exponential overstates variability and therefore overstates queue lengths.
The M/M/1 results everyone quotes — average queue length ρ/(1−ρ), the blow-up as utilisation approaches 1 — depend on both halves. The qualitative lesson survives (queues explode near full utilisation, and that is real), while the specific numbers should be treated as a first approximation.
When service times matter, M/G/1 models allow a general distribution, and the Pollaczek-Khinchine formula shows queue length depends on service-time variance as well as its mean — which is precisely the information an exponential assumption throws away.
Two views of one process
The exponential and the Poisson are the same thing asked differently, and moving between them saves work.
Poisson counts events in a fixed window. How many arrivals this hour?
Exponential measures the gap between them. How long until the next arrival?
They share λ, and each implies the other exactly. The bridge is a single identity: no events in time t means the next gap exceeds t, so P(X > t) = P(Poisson count = 0) = e⁻ᵤᵗ — which is the exponential survival function.
That gives you a free translation. “What is the chance of no calls in the next half hour?” can be answered from either side and must agree.
The generalisation is worth knowing too: the time until the r-th event is gamma distributed, with the exponential as the case r = 1. That mirrors the geometric-to-negative-binomial step exactly, and the parallel is not a coincidence — it is the same structure in continuous and discrete time.
Checking whether your data is exponential
Three checks, cheapest first, and the first two are usually enough.
Compare the standard deviation with the mean. They should be equal — a coefficient of variation of 1. This is a one-line screen and it rules out a great deal. A CV well above 1 means clustering or a mixture of populations; well below 1 means the process is more regular than random.
Compare the median with 0.693 times the mean. A different fingerprint, and it catches departures the CV can miss.
Plot the log of the survival function. For an exponential, ln P(X > t) is exactly −λt — a straight line through the origin. Curvature is the clearest visual evidence the hazard is not constant: bending down means a rising hazard (wear), bending up means a falling one (early failures being weeded out).
That last plot is the most informative because it shows where the model breaks rather than just that it does — the same argument the normal page makes for Q-Q plots over formal normality tests.
If it fails, the usual next step is a Weibull: it nests the exponential, so fitting one and finding a shape parameter near 1 is itself confirmation that the exponential was adequate.
Where it genuinely applies
The pattern is random arrival rather than accumulating wear.
Radioactive decay. The purest case in nature. An atom has no age and no memory, so the hazard is exactly constant. Half-life is precisely the median, ln2/λ.
Arrivals. Customers, requests, packets, calls — whenever they come independently from a large population.
Shock-driven failures. Components that fail from external events rather than degradation: a power surge, a cosmic ray flipping a bit, a pothole. Age genuinely does not matter for those.
The flat middle of the bathtub. Even for things that do wear, the useful-life period between infant mortality and wear-out is well modelled by a constant hazard. Restricting the model to that window is legitimate; extrapolating it past the wear-out knee is not.
The counter-example worth keeping in mind: human lifespans are emphatically not exponential. Mortality rises sharply with age, and a model saying a ninety-year-old is as likely to survive the next year as a twenty-year-old is not slightly wrong. Anything with a biological or mechanical ageing process needs a distribution that permits a changing hazard.
Sources and methodology
References for the distribution and the hazard-rate framing.
Method. Upper tails are computed as e⁻ᵤᵗ directly rather than as 1 − CDF, so a long wait does not lose precision to cancellation. The memorylessness panel evaluates both sides of P(X > s+t | X > s) = P(X > t) from the survival function rather than displaying a rearranged formula, so the identity is demonstrated on the reader’s own numbers. Rate and mean are always both displayed because they are reciprocals and substituting one for the other yields a plausible wrong answer that nothing would otherwise flag. The suite asserts the cross-distribution identity this page rests on — that chi-square with two degrees of freedom is exactly Exponential(1/2) — along with the CDF being monotone and bounded across its range. That engine is verified on every change against 0 assertions. The count and the per-case breakdown are published on the formula verification page.
Related calculators
Where this goes next:
Poisson DistributionPoisson probabilities with a dispersion test against your own variance — because a Poisson forces variance to equal the mean, and real count data usually does not, which is exactly where the tail goes wrong.
Geometric DistributionWaiting until the first success, with both conventions shown — and a panel that computes your remaining wait given past failures and from scratch, proving they are identical. The gambler's fallacy as arithmetic.
Normal DistributionProbabilities under a normal curve in all four directions with the region shaded — and the empirical rule given exactly, because two standard deviations is 95.45% and the 95% everyone quotes sits at 1.96σ.
Binomial DistributionExact binomial probabilities at any n — including thousands, where a factorial overflows — with the normal approximation beside them and its error measured, which is 0.6% at the centre and 261% in the tail.
Hypergeometric DistributionExact probabilities for drawing without replacement, with the binomial answer beside them and the 10% rule evaluated — at a 40% sampling fraction the binomial is 22% wrong, at 0.2% it is 0.06% wrong.
Chi-SquareGoodness of fit and tests of independence with every expected count and per-cell contribution shown — because the validity condition is about expected counts, not observed ones, and most calculators hide them.
An educational tool. The exponential assumes a constant hazard rate — that elapsed age does not affect the chance of an event. For anything subject to wear, fatigue or ageing this is false, and a Weibull or another distribution allowing a changing hazard is appropriate.
Published an exponential distribution calculator that accepts a rate or a mean and always displays both, because they are reciprocals and substituting one for the other produces a plausible-looking wrong answer that nothing would otherwise flag.
Translates memorylessness into its physical claim — a constant hazard rate, meaning a component that has run a thousand hours is as likely to fail in its next hour as a new one — and says plainly that this is false for anything subject to wear, naming the Weibull as the generalisation that allows a changing hazard.
Notes the diagnostic fingerprint: an exponential has standard deviation exactly equal to its mean, so a coefficient of variation far from 1 rules the model out in one line.
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