Answer: P(X > t) - independent of past waiting time.
- A P(X > s), reusing the original condition as the answer
- B P(X > t) - independent of past waiting time
- C P(X > s+t), the unconditional probability of exceeding s+t
- D e<sup>-lambda</sup>, a constant with no dependence on s or t at all
Correct answer: B. P(X > t) - independent of past waiting time
Explanation: Memoryless: P(X > s+t | X > s) = P(X > t). Past waiting time is irrelevant. Only exponential has this property among continuous distributions.
Venn diagram of the universal set U with events A and B, showing the intersection (A and B), the parts unique to each event, and the complement region outside both.
Concept context
Chance, events, conditional probability, and Bayes theorem