Confidence of a failure A joint of steel pipe

Jackson Floyd

Jackson Floyd

Answered question


Confidence of a failure
A joint of steel pipe (casing) has a 1% failure rate. 400 joints of casing are in a typical well. How may wells can I drill or joints can I run, before I am 90% confident of having at least one casing failure? You can leave out the well to simplify the problem: Casing has a 1% failure rate. How many joints of casing can I have before I am 90% confident in having a failure? Thanks for re teaching me. I think it is called gambler's ruin.

Answer & Explanation

Esteban Sloan

Esteban Sloan

Beginner2022-03-28Added 21 answers

This is a problem using the geometric distribution. Distributions with that name are defined in different ways.
My definition is that X is the number of the trial on which we see the first casing failure.
Then the PDF of X is f(k)=P(X=k)=(1p)pk1, for k=1,2,. Also, the CDF is F(k)=P(Xk)=i=1kf(i).
We can use R statistical software to make a table of the PDF and CDF for your problem and then print out the first few and the last few rows of the table:
>x=1:230; pdf=.01.99x1;cdf=cum(pdf)
It seems that it will take 230 fittings to be 90% sure of a casing failure. I will leave it to you to figure out how to sum the geometric series to get the CDF. Then you can solve to find the smallest k such that F(k).9.
Note: The version of the geometric distribution programmed into R is for the random variable Y which counts the trials up to, but not including, the first casing failure. Thus in R, qgeom(.9, .01) returns 229.

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