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If a piece of computer equipment will fail once a year and be unavailable for 1.5 hours, how do I calculate the probability of the failure occurring each hour of every day

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Could you tell us the source of your question? Homework? Work assignment? Idle curiosity? As stated, there seems to be a lot of ambiguity as to what is meant. – Dilip Sarwate Dec 21 '11 at 16:21
How many hours are in a year? Do you count all of them, or only "working" hours of some kind? – Ross Millikan Dec 21 '11 at 16:30
Silly remark, but since it will only fail once a year, it is certain that it won't fail each hour of every day. – Marc van Leeuwen Dec 21 '11 at 18:20

If your question is:

"What is the probability of a failure occurring in a given hour?"

We will count the failure event itself, which happend once per year, giving:
Number of hours in a year = 365d * 24h = 8,760h
Number of failures per hour = 1 failure per year / 8,760h/y = 0.0001142 failures per hour, or:

0.01142% chance of experiencing a failure in a given hour.

If the question was:

"What is the probability of having the computer be unavailable within a given hour?"

(as is implied by the duration given), then:

The probability of the single failure overlapping with a given hour is:

Number of days in a year = 365d/y
Number of days of unavailability = 1.5h / 24h/d = 0.0625d of unavailability per year

0.01712% probability of having some unavailability within a given hour.

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