Probability of Equipment Failure

Question

I have a set of 25 equipment. Each equipment fails (and recover) once a month, on average.

a) What is the probability that 2 equipment will fail on the same day? What is this probability in terms of once in X years?

b) What is the probability that 2 equipment will fail within a 30-minute period? What is this probability in terms of once in X years?

Pere · Accepted Answer · 2018-10-13 09:37:01Z

1

Not an answer but a few hints to start:

What distribution follows the number of failures in a given period? (Hint: search for Poisson).
How does the parameter of that distribution changes when the length of period changes?
With that distribution you can know the probability of a given set falling on a given day. That probability is the same for every set.
Since you have 25 sets, you are performing independent 25 experiments every day, each one with the same probability of success or failure.
Which distribution follows the number of failures in these experiments. (Hint: Your setting is not very different from tossing a coin 25 times).

answered Oct 13, 2018 at 9:37

Pere

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$\begingroup$ I am thinking of approaching it this way: a) For 1 equipment, a failure event will occur once a month (i.e. 30 days) or 0.033 failure per day. Hence, there is a 3.3% chance that an equipment will fail in a day. Since performance of each equipment is independent, the probability of 2 failures within a day can be calculated based on a binominal distribution. Therefore, the probability of exactly 2 failures within a day = 25C2*(0.033)^(2)*(0.966)^(23) = 0.152. $\endgroup$
– Eugene
Commented Oct 13, 2018 at 10:26
$\begingroup$ b) For 1 equipment, a failure event will occur once a month (i.e. 30X48=1,440 30-min periods) or 1/1440 failures per 30-min period. Hence, there is a 0.06944% chance that an equipment will fail in a 30-min period. Since performance of each equipment is independent, the probability of 2 failures within a 30-min period can be calculated based on a binominal distribution. Therefore, the probability of exactly 2 failures within a 30-min period = 25C2*(0.0006944)^(2)*(0.99930556)^(23) = 0.000142365. If this approach is correct, how do I translate these probabilities to once in X years? $\endgroup$
– Eugene
Commented Oct 13, 2018 at 10:26
1

$\begingroup$ I suggest understanding the Poisson distribution before attempting the start of the problem, although since the conditions on your problem are near to those that allow approximation of Poisson and binomial your numbers may be not very far off. $\endgroup$
– Pere
Commented Oct 13, 2018 at 10:39

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