Hours until one of two alarms first rings

Question

Two independent alarms are wired to the same room. Each hour, alarm A rings with probability $1/2$ and alarm B rings with probability $1/3$ , independently of each other and across hours. You wait until you hear at least one of them ring.

On average, how many hours until the first ring from either alarm?

Accepted Answer

Waiting for the first of several recurring events is still a wait-for-first-success problem, once you find the probability that something happens in a given hour. In any single hour, the chance that neither alarm rings is P(	ext{no ring}) = \left(1 - 	frac12ight)\left(1 - 	frac13ight) = 	frac12 \cdot 	frac23 = 	frac13. So the chance at least one rings is p = 1 - 	frac13 = 	frac23. Each hour is then an independent trial with success probability p, and the expected wait is 1/p: E = \frac{1}{p} = \frac{1}{2/3} = \frac{3}{2} = 1.5 	ext{ hours.} Takeaway. For the first of independent …

Hours until one of two alarms first rings

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