Worked Exponential Waiting-Time Problems

Introduction

A waiting-time question often combines several steps: interpret an event rate, express it in the same time unit as the thresholds, choose the appropriate probability formula, and explain what the result means. Keeping those steps separate helps avoid errors that can be hard to spot in a final numerical answer.

Throughout this tutorial, a waiting time \(T\) is modeled as exponential with rate \(\lambda>0\), measured in events per unit of time. This model is appropriate when the waiting time is assumed to have a constant event rate over time. It is not appropriate when the rate changes over the waiting period or when the problem specifies a different waiting-time model.

Definition: If \(T\sim\operatorname{Exponential}(\lambda)\), then \(T\) is a nonnegative waiting time with survival function \(P(T>t)=e^{-\lambda t}\) for \(t\geq 0\), cumulative distribution function \(P(T\leq t)=1-e^{-\lambda t}\), and mean \(E[T]=1/\lambda\). The rate \(\lambda\) and time \(t\) must use compatible units, so that \(\lambda t\) is dimensionless.

The survival function gives the probability that the wait exceeds a threshold. Subtracting two survival probabilities gives the probability that the wait falls between two thresholds. The mean is a long-run average waiting time under the model; it is not a promise that an event will occur at that time.

A Reliable Calculation Sequence

For multi-step problems, first write down the rate and its time unit. If the thresholds use another time unit, convert either the rate or the thresholds before substituting into an exponential. Then identify whether the question asks for a probability before a threshold, after a threshold, within an interval, or conditional on an earlier wait.

State the model and rate.
Write \(T\sim\operatorname{Exponential}(\lambda)\), including the units of \(\lambda\), and state the assumption that the rate is constant.

Align the units.
Express each threshold in the rate's time unit, or convert the rate to the thresholds' unit.

Choose the probability expression.
Use \(1-e^{-\lambda t}\) for \(P(T\leq t)\), \(e^{-\lambda t}\) for \(P(T>t)\), or a difference of survival probabilities for an interval.

Interpret and check.
State the result in context, round only at the end, and check that probabilities lie between 0 and 1 and have the requested event direction.

Threshold and Interval Probabilities

For \(0\leq a<b\), an interval probability is found by subtracting the probability of waiting longer than \(b\) from the probability of waiting longer than \(a\). The event \(\{T>a\}\) includes every wait that exceeds \(b\), as well as waits between \(a\) and \(b\). Removing the longer waits leaves the desired interval.

P(a<T\leq b)=P(T>a)-P(T>b) =e^{-\lambda a}-e^{-\lambda b}.

Because an exponential distribution is continuous, the probability of waiting exactly \(a\) or exactly \(b\) is zero. Thus the probability is unchanged if either interval endpoint is included instead of excluded. This endpoint fact does not mean that a threshold can be converted to a different unit without also converting the rate.

Worked Example: A Service Interruption at Two Thresholds

A hypothetical pump has an exponentially distributed time \(T\) to its next service interruption, with mean 4 hours. Assume its interruption rate is constant. Find (1) the probability of an interruption within 2 hours and (2) the probability that the wait is between 2 and 5 hours. Give probabilities to four decimal places.

Method and model. For an exponential waiting time, \(\lambda=1/E[T]\). The mean is 4 hours, so

\lambda=\frac{1}{4}=0.25\text{ per hour},\qquad T\sim\operatorname{Exponential}(0.25).

(1) Within 2 hours. “Within” means \(T\leq 2\). The threshold is already in hours, so

P(T\leq 2)=1-e^{-(0.25)(2)} =1-e^{-0.5} =1-0.6065307 =0.3934693.

Rounded to four decimal places, the probability is 0.3935.

(2) Between 2 and 5 hours. Apply the interval formula with \(a=2\) and \(b=5\):

\begin{aligned} P(2<T\leq 5) &=e^{-(0.25)(2)}-e^{-(0.25)(5)}\\ &=e^{-0.5}-e^{-1.25}\\ &=0.6065307-0.2865048\\ &=0.3200259. \end{aligned}

Rounded to four decimal places, this probability is 0.3200. The result is positive and smaller than \(P(T>2)=0.6065\), as it must be because the interval event is only part of the event \(T>2\).

Common mistake avoided: Subtract the two survival probabilities using the same rate and units. Here \(0.6065307-0.2865048=0.3200259\), which rounds to 0.3200, not 0.3205.

Converting Rates Before Calculating

A rate may be reported per hour while the thresholds are in minutes, or per day while the thresholds are in hours. Both representations are acceptable if used consistently. For example, a rate of 12 events per hour is \(12/60=0.2\) events per minute. Equivalently, a threshold of 2 minutes is \(2/60\) hours, and the product \(12(2/60)\) is the same as \(0.2(2)\).

A useful check is to inspect the exponent: its rate-times-time product must be unitless. Multiplying an hourly rate directly by a number of minutes is not a valid exponential calculation.

Worked Example: A Request Arrives After an Elapsed Wait

A hypothetical support line receives requests at a constant rate of 12 per hour. Given that no request has arrived during the first 5 minutes, find the probability that the next request arrives between 2 and 7 additional minutes from now. Give the probability to four decimal places.

Method and model. Let \(T\) be the time from the start of observation to the next request. Under the constant-rate exponential model, \(T\sim\operatorname{Exponential}(12)\) when time is measured in hours. The question concerns the additional wait after the first 5 minutes with no request. The memoryless property of the exponential distribution says that, conditional on \(T>5\) minutes, the remaining wait has the same exponential distribution as a new wait.

To see the basis for this conditional result, for \(s,t\geq0\) the survival function gives

P(T>s+t\mid T>s) =\frac{P(T>s+t)}{P(T>s)} =\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} =e^{-\lambda t}.

The right side is the survival probability for a fresh exponential wait of length \(t\). Thus the additional wait has rate 12 per hour, or \(12/60=0.2\) per minute. Let \(W\) denote this additional wait, in minutes. The requested event is \(2<W\leq7\), so

\begin{aligned} P(2<W\leq7) &=e^{-(0.2)(2)}-e^{-(0.2)(7)}\\ &=e^{-0.4}-e^{-1.4}\\ &=0.6703200-0.2465970\\ &=0.4237230. \end{aligned}

Rounded to four decimal places, the probability is 0.4237. The original 5-minute wait does not get added to the requested 2- and 7-minute additional-wait thresholds; those thresholds are explicitly measured from now.

Common mistake avoided: Do not calculate the probability that the total wait from the start lies between 2 and 7 minutes. The condition says 5 minutes have already passed, and the question asks about an additional wait.

Rate, Mean, and Time-Unit Interpretation

The exponential mean is the reciprocal of the rate only when the rate and mean use reciprocal units. If \(\lambda\) is in events per hour, \(1/\lambda\) is in hours. A reported event rate such as 9 per day can be converted to an hourly rate by dividing by 24. It can also be left per day if every time threshold is expressed in days.

Worked Example: Daily Rate, Hourly Thresholds, and Mean Wait

A hypothetical monitoring system records events at a constant rate of 9 per day. Let \(T\) be the waiting time to the next event. Find (1) the mean wait, (2) the probability the wait exceeds 2 hours, and (3) the probability that the wait is between 1 and 4 hours. Give probabilities to four decimal places.

Method and model. Convert the rate to events per hour so it matches the thresholds:

\lambda=\frac{9}{24}=0.375\text{ per hour},\qquad T\sim\operatorname{Exponential}(0.375).

(1) Mean wait. Taking the reciprocal of the hourly rate gives

E[T]=\frac{1}{0.375}=\frac{8}{3}\text{ hours} =2\frac{2}{3}\text{ hours} =2\text{ hours }40\text{ minutes}.

(2) More than 2 hours. Use the survival function:

P(T>2)=e^{-(0.375)(2)} =e^{-0.75} =0.4723666 \approx \mathbf{0.4724}.

(3) Between 1 and 4 hours. Subtract the survival probability at 4 hours from that at 1 hour:

\begin{aligned} P(1<T\leq4) &=e^{-(0.375)(1)}-e^{-(0.375)(4)}\\ &=e^{-0.375}-e^{-1.5}\\ &=0.6872893-0.2231302\\ &=0.4641591 \approx \mathbf{0.4642}. \end{aligned}

The interval probability is less than \(P(T>1)=0.6873\), which is a useful check because the interval is contained within waits exceeding 1 hour.

Common mistake avoided: Using \(1/9\) as a mean in hours would confuse a per-day rate with an hourly rate. In days the mean is \(1/9\) day; converting gives \((1/9)(24)=8/3\) hours, the same answer.

Conditional Waiting Times and Exam Traps

When a problem says no event has occurred by time \(s\), distinguish carefully between a total waiting time and an additional waiting time. Under an exponential model, the conditional additional wait has the same rate as the original wait. If the question instead asks for the total wait to exceed \(s+t\), the event is \(\{T>s+t\}\); if it asks for the additional wait to exceed \(t\), the relevant conditional event is \(\{T>s+t\mid T>s\}\). These are related, but they are not the same probability.

Common checks for a multi-step problem include the following:

Check the rate units. Convert per day, per hour, or per minute before multiplying by a threshold.
Check the event wording. “Within \(t\)” uses \(1-e^{-\lambda t}\), whereas “more than \(t\)” uses \(e^{-\lambda t}\).
Check the interval order. For \(a<b\), \(e^{-\lambda a}\) must be greater than \(e^{-\lambda b}\), so their difference is nonnegative.
Check the conditional clock. If thresholds are measured from now after an elapsed wait, use the additional-wait thresholds, not the original elapsed time plus those thresholds.
Check the model assumption. The exponential formulas require an exponential waiting-time model, with a constant rate over the period considered.

Key takeaway: Write the rate with its units, align those units with every threshold, and identify whether the question concerns a threshold, an interval, or a conditional additional wait. Then apply the matching survival-function formula and round only after completing the calculation.

Check Your Understanding

For each problem, state the model and keep the rate and time units consistent. Probabilities in the answers are rounded to four decimal places.

A hypothetical inspection station handles 6 items per 8-hour shift at a constant rate. Let \(T\) be the wait in hours to the next item. Find \(P(T>1.5)\) and \(P(1<T\leq2)\). (Answer: \(0.3247\); \(0.2492\))
A hypothetical alert has an exponential waiting time with mean 10 minutes. Find the probability the wait exceeds 15 minutes and the probability it is between 5 and 20 minutes. (Answer: \(0.2231\); \(0.4712\))
A hypothetical queue receives entries at a constant rate of 15 per hour. Given that no entry has arrived during the first 4 minutes, find the probability the next entry arrives between 1 and 3 additional minutes from now. (Answer: \(0.3064\))
A hypothetical sensor registers events at a constant rate of 2.4 per hour. Find the probability of at least one event within 10 minutes, and find the mean waiting time in minutes. (Answer: \(0.3297\); 25 minutes)
A hypothetical processing station completes 5 jobs every 2 hours at a constant rate. Given that no job has completed during the first 20 minutes, find the probability that the additional wait exceeds 45 minutes. (Answer: \(0.1534\))

What You'll Learn