The exponential distribution describes the amount of time we wait between random events that occur at a constant average rate. It is most commonly used to model the time until the next event in a Poisson process.
f(x) = λ e−λx for x ≥ 0F(x) = 1 − e−λx1 / λ    Standard Deviation (σ): 1 / λIf something happens on average 5 times per hour (λ = 5/hour), then the average time between events is 1/5 hour = 12 minutes.
The exponential distribution is unique because it has no memory. The chance of the next event occurring in the next x minutes is the same—no matter how long you’ve already waited.
Formula: P(X < t₁ + t₂ | X > t₁) = P(X < t₂)
The exponential distribution models the time between events in a Poisson process. Its probability density function (PDF) is:
f(x) = λ e−λx, for x ≥ 0
Mean: E(X) = 1/λ Variance: Var(X) = 1/λ²
Problem: Log-ons occur at a rate of 25 per hour. What is the probability of waiting more than 6 minutes (0.1 hour) for the next log-on?
P(X > 0.1).P(X > x) = 1 − P(X ≤ x).F(x) = 1 − e−λx, so
        P(X > 0.1) = 1 − P(X ≤ 0.1)= 1 − F(0.1)= 1 − (1 − e−λ·0.1)= e−λ·0.1 = e−25·0.1 = e−2.5 ≈ 0.08208This uses the complement of the CDF (survival): P(X > x) = e−λx. The PDF is the curve’s height, not a probability by itself.
Property: P(X < t₁+t₂ | X > t₁) = P(X < t₂)
1 − e−0.714×0.5 ≈ 0.30.This is why the exponential is called “memoryless.”
Scenario: On average, the Wi-Fi drops once every 20 minutes. What’s the probability it stays up for a 10-minute video?
⬇️ Download: Exponential_Probability_Calculator.xlsx
| Quantity | Excel formula (generic) | With your numbers | Value | 
|---|---|---|---|
| Mean wait (1/λ) | =1/lambda |  |  | 
| Variance (1/λ²) | =1/lambda^2 |  |  | 
| P(X > x) (survival) | =EXP(-lambda*x)= 1 − (1 − EXP(−λx))  (shows complement step) |  |  | 
| P(X ≤ x) (cdf) | =1-EXP(-lambda*x) |  |  | 
| P(a < X < b) | =EXP(-lambda*a)-EXP(-lambda*b)= (1−EXP(−λb)) − (1−EXP(−λa)) |  |  | 
| Poisson link check: P(N(x)=0) | =EXP(-lambda*x) |  |  | 
λ in B5, x in B6, a in B7, b in B8=EXP(-B5*B6) and (complement) =1-(1-EXP(-B5*B6))=1-EXP(-B5*B6)=EXP(-B5*B7)-EXP(-B5*B8)=1/B5 ; Variance: =1/B5^2=EXP(-B5*B6)In a Poisson process with rate λ, the number of events in a window of length t is Poisson with mean λt. The probability of zero events in that window equals the probability that the waiting time to the first event exceeds t: P(N(t)=0) = e^{−λt} = P(T > t)