You compute probabilities for waiting times. One wait at a constant rate follows the exponential distribution. Its mean is 1 divided by the rate. The total of several independent waits follows the gamma distribution. A problem belongs here when it gives a waiting time with a stated mean or rate and asks for the chance of exceeding a time, finishing by a time, or waiting longer after some time has already passed. You will practice: exponential survival function, memorylessness, Erlang/gamma sums, gamma moments.
Packets arrive at a network router so that the time between consecutive packet arrivals is exponentially distributed with mean 2 seconds. A packet has just arrived. What is the probability that the router waits more than 4 seconds for the next packet? Give your answer to 4 decimal places.
Answer:
The interarrival time X is exponential with mean 2 seconds, so the rate is 1/2 per second.
The next gap exceeds 4 seconds with the exponential tail probability.
In a bacterial culture, the waiting time until a particular mutation first appears is exponentially distributed with mean 5 hours. The culture has been observed for 3 hours with no mutation yet. What is the probability that no mutation appears during the next 4 hours? Give your answer to 4 decimal places.
Answer:
By memorylessness, the 3 hours already observed do not matter.
The remaining wait is a fresh exponential with rate 1/5 per hour.
In a soccer match, the waiting time from kickoff until the first goal is exponentially distributed with mean 30 minutes. What is the probability that the first half (45 minutes) ends with no goal having been scored?
Answer:
Mean 30 minutes gives rate 1/30 per minute.
No goal in the first half means the exponential wait exceeds 45 minutes.
The length L of a customer-support phone call is exponentially distributed with mean 10 minutes. Compute E[L²] (in minutes²).
Answer:
For an exponential, the variance is the square of the mean, so it is 100 here and the mean is 10.
At a national park, the waiting time between eruptions of a geyser is exponentially distributed with mean 20 minutes. A hiker arrives just as an eruption ends. What is the probability that she must wait more than 40 minutes for the next eruption?
Answer:
Use the tail of an exponential with mean 20 and plug in 40 minutes.
The lifetime of an LED module in a streetlight is exponentially distributed with mean 6 years. Compute the variance of the lifetime (in years²).
Answer:
For an exponential with mean m, the second moment is 2 times m squared, so the variance is m squared.
You are stargazing during a meteor shower. The waiting time until the next shooting star is exponentially distributed with mean 4 minutes. What is the probability that you wait more than 4 minutes?
Answer:
Use the exponential tail P(T > t) = e^(-t/mean).
A firefly flashes at random times. The waiting time until its next flash is exponentially distributed with mean 5 seconds. What is the probability that it flashes within the next 10 seconds?
Answer:
Take the complement of the tail.
A Geiger counter registers clicks so that the time until the next click is exponentially distributed with rate 0.2 clicks per second. What is the probability that there is no click during the next 10 seconds?
Answer:
With rate given, the tail is P(T > t) = e^(-λt).
The time until a fisherman gets a bite is exponentially distributed with mean 15 minutes. He has already waited 30 minutes without a bite. What is the probability that he gets a bite within the next 15 minutes?
Answer:
By memorylessness the 30 minutes already waited do not matter; the remaining wait is a fresh exponential with mean 15.
A robotic arm welds 4 joints one after another. Each weld takes an independent exponentially distributed time with mean 6 minutes. Let S be the total welding time. Compute Var(S) in minutes².
Answer:
S is Erlang: a sum of 4 independent exponentials. Each has variance mean² = 36, and independent variances add.
The burn time T of a scented candle is exponentially distributed with mean 3 hours. Compute E[T²] in hours².
Answer:
For an exponential, E[T²] = 2/λ² = 2·mean².
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