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So I am an engineer. We have a method at our organization of defining the probability of an equipment failure based on frequency of occurrence. So for example we may categorize a piece of equipment failing falling under a category of failing once in between 1 to 10 years or 10 to 100 years, or even less than 1 year for high risk items.

I felt this fit well into a Poisson Distribution to do so in order to determine a percentage for the probability of exactly n of these independent random variables of occurring once per year.

$$

\begin{array}{llllll} Machinery & \text{Frequency of Occurance} & \text{Assumption Frequecy per Year (Expected Value)} & \text{Probability via Poisson Distribution of Occuring Once Per Year Pr(k=1)} & \text{Probability via Poisson Distribution of Occuring Once or More Per Year (Pr(k>=1)} & \text{Probability via Poisson Distribution of Not Ocuuring Per Year Pr(k=0)} \\ A & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ B & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ C & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ D & \text{1/yr to 1/10yr} & 0.20 & 0.163746150615596000 & 0.017523096306421800 & 0.818730753077982 \\ E & \text{1/yr to 1/10yr} & 0.20 & 0.163746150615596000 & 0.017523096306421800 & 0.818730753077982 \\ F & \text{1/yr to 1/10yr} & 0.20 & 0.163746150615596000 & 0.017523096306421800 & 0.818730753077982 \\ G & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ H & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ I & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ J & \text{1/10yr to 1/100yr} & 0.02 & 0.019603973466135100 & 0.000197353227109676 & 0.980198673306755 \\ K & \text{1/100yr to 1/1000yr} & 0.002 & 0.001996003997334670 & 0.000001997335332238 & 0.998001998667333 \\ L & \text{1/100yr to 1/1000yr} & 0.002 & 0.001996003997334670 & 0.000001997335332238 & 0.998001998667333 \\ M & \text{1/100yr to 1/1000yr} & 0.002 & 0.001996003997334670 & 0.000001997335332238 & 0.998001998667333 \\ N & \text{1/yr to 1/10yr} & 0.20 & 0.163746150615596000 & 0.017523096306421800 & 0.818730753077982 \\ O & \text{1/100yr to 1/1000yr} & 0.002 & 0.001996003997334670 & 0.000001997335332238 & 0.998001998667333 \end{array} $$

Can someone check where logic wrong with my math? I don't think I am right because this does not converge to 1 as the sample size increases to infinity? Where X = Event occurring once in a year and X' = Event not occurring in a year.

$$ \Pr\left[\text{One event occurring per year} \right] = \Pr \left[ A | B' \cap C' \cap D' \cap...\right] + \Pr \left[ B | A' \cap C' \cap D' \cap...\right]+ \Pr \left[ C | A' \cap B' \cap D' \cap...\right]+ ...=\frac{\Pr\left[A \cap B' \cap C' \cap D' \cap... \right]}{\Pr\left[ A \right]}+\frac{\Pr\left[ B \cap A' \cap C' \cap D' \cap...\right]}{\Pr\left[ B\right]}+\frac{\Pr\left[C \cap A' \cap B' \cap D' \cap... \right]}{\Pr\left[ C\right]}+...= \Pr[B']\cdot\Pr[C']\cdot Pr[D']\cdot...+\Pr[B']\cdot\Pr[C']\cdot Pr[D']\cdot...+\Pr[A']\cdot\Pr[B']\cdot Pr[D']\cdot...+... $$

I am looking for the probability that exactly n events occur in a year. An event would be any equipment failure. Realistically, the event would only occur once per piece of equipment. I did use the poisson distribution for calculating probabilities for each piece of equipment as shown by my chart. I wondering as an overview for all pieces of equipment am I able to determine a probability of any one piece of equipment failing per year? ... any two failing per year?

I was a stats major many moons ago and I am employed now as a mechanical engineer. So my logic is probably off here.

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    $\begingroup$ Your question is a little unclear: are you asking for the chance that exactly $n$ events occur in a year or that at least $n$ events occur? And would an event be "any equipment failure" or would it be "one or more failures of a given machine," so that $n$ counts the number of machines that fail, regardless of how many times each fails? And since you state you are using a Poisson distribution, why not use the formulas for its probabilities? $\endgroup$
    – whuber
    Mar 4, 2021 at 21:38
  • $\begingroup$ I am looking for the probability that exactly n events occur in a year. An event would be any equipment failure. Realistically, the event would only occur once per piece of equipment. I did use the poisson distribution for calculating probabilities for each piece of equipment as shown by my chart. I wondering as an overview for all pieces of equipment am I able to determine a probability of any one machine failing per year? ... any two failing per year? $\endgroup$ Mar 8, 2021 at 23:08
  • $\begingroup$ math.stackexchange.com/questions/396214/… I believe I found what I was looking for on this question. $\endgroup$ Mar 12, 2021 at 5:38

1 Answer 1

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You can model each piece of equipment using exponential distribution. Then, the probability that this piece of equipment will fail after time $t$ is

$$p_i(t) = 1 - \exp(- t / \tau_i)$$

where $\tau_i$ is the expected fail time for this piece of equipment. You seem to be interested $t = t_0 = 1$ year. Compute probabilities via

$$P_i = p_i(t_0)$$

Next, we want to know the probability that exactly $n$ out of $N$ events will occur. To do so, we need to go over all ways to select $n$ events out of $N$, compute the probability of occurence for each of the combinations, and then sum them up. Let $s$ be a vector of 1s and 0s, indicating which pieces of equipment failed. We require that $\sum_i s_i = n$. Then if, we know $s$, the probability of that situation is $$P[s] = \prod_{i=1}^N P_i^{s_i} (1 - P_i)^{1-s_i}$$ That is, we multiply the result by $P_i$ if $s_i=1$, otherwise we multiply the result by $1 - P_i$. Finally, to find the probability of exactly $N$ events, we need to sum $P[s]$ over all combinations $s \in S(n, N)$.

$$P[N] = \sum_{s \in S(n, N)} P[s]$$

If $N$ is not too large (10-20), you may be able to write a brute-force code to compute $P[N]$ and get an answer within a few seconds. However, if $N >> 20$, you will need to optimize the code for computing the result, so that you have a chance of getting the answer within the age of the universe :D. I can provide a python code for this solution if you feel that it is necessary.

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