# Calculate probability of a particular event

I want to calculate the probability of an outage occurring on a particular transmission tower given the following information.

    tower_type  outages_tower_type  total_towers_of_type
a           51                  391
b           64                  1172
c           89                  13580
d           17                  2094
e           13                  797

total towers:18034
time period: 5 years


Can you also give me suggestions on other ways to calculate the probability in this situation? I'm familiar with Poisson distribution for instance.

• Just a clarification, when you say on a particular transmission tower, do you mean the probability that a tower of type $x$ has an outage, or that any tower of any type has an outage? – Emil Aug 22 '18 at 8:40
• probability that a tower of type x has an outage – user1361488 Aug 22 '18 at 8:55
• You mean the probability within a declared time interval? – Denziloe Aug 22 '18 at 14:02

Let's try to find the probability that type A towers (collectively) have at least one outage within the next month.

With 51 outages over 5 years (60 months) the estimated monthly outage rate for type A towers is $\hat \lambda = 51/60 = 0.85.$ If we assume the outage rate is constant and can be used for future months, then the number $X$ of outages in the next month has $X \sim \mathsf{Pois}(0.85).$ Thus $$P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-0.85} = 0.5726.$$

1 - exp(-0.85)
[1] 0.5725851
1 - dpois(0, 0.85)
[1] 0.5725851


(Computations use R.)

The value $\hat \lambda$ for the monthly outage rate is an estimate. A 95% confidence interval for the five-year outage rate $\lambda_{60}$ is of the type $$X_{60}+2 \pm 1.96\sqrt{X_{60} + 1},$$ which computes to $(38.87, 67.13).$ Thus a confidence interval for the monthly rate is $(0.648, 1.119).$

The best guess is that the probability of a type A tower outage in the next month is 0.5726 (as above), but that probability was based on an estimate. Using the upper confidence limit for $\lambda,$ we can be reasonably confident that the actual probability is below $0.6734.$

53 + c(-1,1)*1.96*sqrt(52)
[1] 38.86624 67.13376
(53 + c(-1,1)*1.96*sqrt(52))/60
[1] 0.6477707 1.1188960
1 - dpois(0, 1.119)
[1] 0.6733938


Note: There is discussion what style of CI to use for the Poisson rate. Two other possibilities are:

(a) The traditional Wald interval $X \pm 1.96\sqrt{X},$ which is known to have bad coverage properties (especially for small $X).$ For type A it gives the interval $(0.617, 1.083)$.

(b) A frequentist interval based on a Bayesian argument with an 'improper' (non-informative) prior uses quantiles .025 and .975 of $\mathsf{Gamma}(51,1).$ Its results are often similar to the interval I used above in the main part of my answer. For type A it gives $(0.633, 1.099)$.

(51 + c(-1,1)*1.96*sqrt(51))/60
[1] 0.6167133 1.0832867
qgamma(c(.025,.975), 51, 1)/60
[1] 0.6328808 1.0986461

• Hi, thanks for the response. How can you account for the total amount of towers per type in a probability calculation? – user1361488 Aug 24 '18 at 9:33
• I didn't. I thought 51 outages was for all 395 type A towers. If 51 is avg per tower, then my computation applies to one type A tower. Always with Poisson computations you need to scale $\lambda$ to match both number of towers and length of time. Once you have the correct $\lambda$ you can follow the pattern of the rest of my computations. – BruceET Aug 24 '18 at 9:41
• ok, as another scenario, replace towers with cars and outages with stolen. Then for instance, how can I determine the probability that my particular type of car would be stolen in the next month based on the amount already stolen and the amount sold in 5 years. – user1361488 Aug 24 '18 at 11:32
• Except for the numbers, how would that be different, if you assume auto thefts to be Poisson with a stable rate that would apply to your car? Is there some part of it you don't know how to do? [Maybe car theft is a flawed example: Across a large city with diverse neighborhoods, theft rates can vary every few blocks -- and depending on whether the car is in your locked garage, in the driveway in front of your garage, in an attended parking lot, or parked on the street with the key left in the ignition.] – BruceET Aug 24 '18 at 17:53