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April 4th, 2019, 12:47 PM   #1
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Probability of 3 or more sample failures on a rolling 12 month basis

Hi everyone, i need some help with the below problem please.

A factory takes monthly samples for quality checks and the chance of passing an individual sample is 95%. The factory is prosecuted if it picks up 3 or more fails in a calendar year. What is the probability of
a) the factory picking up 3 or more failures within a single calendar year (i.e. after 12 samples).
b) the chance of the factory being prosecuted one or more times over the course of 10 years.

The regulary regime in the factory then changes such that if the factory picks up 3 failures on any ROLLING 12 period (i.e. Feb-Jan, Mar-Feb etc) it can be prosecuted.

c) what is the probability that within any calendar year the factory could be liable to prosecution i.e. has had 3 or more fails on its books at any one time over the course of 12 samples.

My answers so far....

a) the odds of this i’ve worked out to be 1.96% using standard binomial cumulative probability. So OK here I think.
b). I used the probability of the factory being prosecuted in any one year (1.96%) and again applied binomial cumulative probability to give a result of 17.9%. Again, i think i’m ok here.
c) This is where i’m stuck. At first i tried to treat the 12 periods (Feb-Jan, Mar-Apr etc) as independent events, like i did for question b), thinking of each event with a probability of 1.96% of being prosecuted. On this basis i calculated the probability to be 21.1%. HOWEVER, when i worked out the probability of having 3 or more fails within 23 months (as a check, reasoning that the probability of this must be AT LEAST 21.1%, as 23 months is the full window of time we’re considering), the result was only 10.52%. On this basis i would expect the probability to be less than 10.52%, as we’re interested only in thoses instances where the 3 failures are clustered within 12 month periods. Any help on this would be greatly appreciated.
Josefk is offline  
April 18th, 2019, 05:10 PM   #2
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Here's why I think the independent events method that produced 21.1% is wrong. If the first 12 months have 0 or 1 fail, that guarantees that there will not be 3 fails in any 12 consecutive months within the first 13 months.
EvanJ is offline  
April 19th, 2019, 05:15 PM   #3
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Math Focus: primes of course
noting the calendar year wording,
c) presuming the threshold was not already met ,
equivalent to chance of
had 2 fails prior 11 mos then new fail Jan
had 1 or 2 fail prior 10 mos then new fail(s) in first 2 calendar mos
meaning P(1 fail in 10 mos) plus 2 consec fails or P(2 fail in 10 mos) plus fail in one of next 2 mos
had 1 or 2 fail prior 9 mos then new fail(s) in first 3 calendar mos
had 1 or 2 fail prior 8 mos then new fail(s) in first 4 calendar mos
(3) new fails over 12 new months

All of those values can be calculated
billymac00 is offline  

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