A radioactive atom has been observed for 30 minutes without decaying. Compared to a freshly observed atom, what is the probability that it decays in the next 10 minutes?
AHigher — the atom is 'overdue' for decay after surviving so long
BLower — having already survived 30 minutes suggests it is unusually stable
CExactly the same — the exponential distribution is memoryless, so elapsed time gives no information about future waiting time
DIt depends on the specific decay rate λ of the element
The memoryless property states P(X > s + t | X > s) = P(X > t). The probability of surviving at least t more minutes, given you have already survived s minutes, is identical to the probability of surviving t minutes from the start. The elapsed time s carries no information. Options A and B both assume the atom has some kind of 'memory' of its past — this is precisely what the memoryless property rules out. This is actually the defining property of the exponential distribution: it is the only continuous distribution with this characteristic.
Question 2 Multiple Choice
Events occur according to a Poisson process at a rate of 4 per hour. What is the mean waiting time between successive events?
A4 hours
B0.25 hours (15 minutes)
C16 hours
D2 hours
If events occur at rate λ per unit time, the inter-arrival times follow an exponential distribution with the same rate parameter λ. The mean of Exp(λ) is 1/λ. With λ = 4 events per hour, the mean waiting time is 1/4 hour = 15 minutes = 0.25 hours. This is one of the most common confusions: λ is the rate (events per time unit), not the mean waiting time. The mean waiting time is 1/λ — the inverse. A higher rate means shorter waits, not longer ones.
Question 3 True / False
You have been waiting 20 minutes for a bus whose inter-arrival times follow an exponential distribution. Your expected remaining wait time is less than 20 minutes, because you are statistically due for a bus soon.
TTrue
FFalse
Answer: False
This is the gambler's fallacy applied to the exponential distribution, and the memoryless property directly refutes it. Having already waited 20 minutes gives you no information about when the next bus will arrive. Your remaining wait time has exactly the same distribution as the full wait of someone who just arrived at the stop. If the mean inter-arrival time is 20 minutes, your expected remaining wait is still 20 minutes — not less. The past waiting time is statistically irrelevant.
Question 4 True / False
The exponential distribution is the continuous analog of the geometric distribution.
TTrue
FFalse
Answer: True
Both distributions model 'waiting time until the first success' and both are memoryless — but in different settings. The geometric distribution is discrete: it models the number of Bernoulli trials until the first success. The exponential distribution is continuous: it models the time until the first event in a Poisson process. Both share the memoryless property (the only distributions with this property in their respective domains), and both can be derived as limiting cases of each other as the time step shrinks to zero. This parallel structure is a key reason the exponential distribution has such a central role in continuous probability.
Question 5 Short Answer
Explain the memoryless property of the exponential distribution in plain language, give a real-world example where it is an appropriate model, and give one where it is not.
Think about your answer, then reveal below.
Model answer: The memoryless property means that the probability of waiting at least t more time units is the same regardless of how long you have already waited. Past elapsed time gives no information about future waiting time. Appropriate example: radioactive decay — an atom is equally likely to decay in the next second whether it was created one second ago or one million years ago (no physical 'aging' mechanism). Inappropriate example: machine failure — a machine is typically more likely to fail the older it gets (wear accumulates over time). The Weibull distribution, not the exponential, is appropriate for aging systems because it allows a hazard rate that increases over time.
The memoryless property holds because the survival function of the exponential is a pure exponential: P(X > x) = e^(−λx). This means P(X > s + t) = P(X > s) × P(X > t) — the probability of surviving past s+t factors exactly as if the two intervals were independent fresh trials. Any distribution with this property must be exponential (by a uniqueness theorem). The practical test: does elapsed time change the 'hazard rate'? If yes, use a different distribution.