What Are the Probabilities of Drilling Outcomes for an Oil Company?

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Homework Help Overview

The discussion revolves around the probabilities associated with drilling outcomes for an oil company, specifically focusing on the classification of wells as either producer or dry wells. The context involves a National Oil Company conducting exploratory drilling operations in the southwestern United States, with a known probability of 15% for a well to be a producer well.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of binomial distribution to calculate the probabilities of different outcomes for 12 wells, including scenarios where all wells are producer wells, all are dry wells, and exactly one is a producer well. Questions arise regarding the accuracy of the calculations and the interpretation of probabilities.

Discussion Status

Some participants provide feedback on the calculations presented, suggesting clarifications on notation and numerical representation. There is an ongoing exchange of ideas regarding the interpretation of the results and the proper application of statistical methods, with no explicit consensus reached on the interpretations.

Contextual Notes

Participants note potential errors in notation and the need for clearer communication of statistical terms. There is also mention of forum rules regarding posting new questions and sharing work, indicating a structured approach to the discussion.

ORACLE
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Questions
1. National Oil Company conducts exploratory oil drilling operations in the south western United States. To fund the operation, investors form partnerships, which provide the financial support necessary to drill a fixed number of oil wells. Each well drilled is classified as a producer well or a dry well. Past experience shows that 15% of all wells drilled are producer wells. A newly formed partnership has provided the financial support for drilling at 12 exploratory locations.

1.What is the probability that all 12 wells will be producer wells?

We have to find the probability of all the 12 wells drilled are producer wells with a 15% prior probability. Here we will calculate the probability using binomial distribution method.
The formula is
P(k out of n) =[n!/k!(n-k)! ](p^k)(q^n-k)
where k =12, number of times a producer well drilled, p = 15% or .15 is the observed probability of a producer well , q = 85% or .85 is the complementary probability (1-p) that is of a dry well, and n = 12 is the number of wells drilled.
Substituted as
P(12 12) =[479001600/479001600(1)](.1512)(.851)
P=0.000000000129746337890625
We can observe that the chances of finding all 12 wells as producer wells are approximately nill.

2.What is the probability that all 12 wells will be dry wells?

Here, we will find the probability of all the 12 wells drilled are dry wells with 85% prior probability.
k =12, number of times a dry well drilled, p = 85% or .85 is the observed probability of a dry well , q = 15% or .15 is the complementary probability (1-p) that is of a producer well, and n = 12 is the number of wells drilled.
Substituted as
P(12 out of 12) =[479001600/479001600(1)](.8512)(.151)
P=0.142241757136172119140625
We can observe a 14.2% chance of finding all 12 wells as dry wells.


3.What is the probability that exactly one well will be a producer well?

k =1, number of times a producer well drilled, p = 15% or .15 is the observed probability of a producer well , q = 85% or .85 is the complementary probability (1-p) that is of a dry well, and n = 12 is the number of wells drilled.
Substituted as
P(1/12) =[479001600/1(39916800)](.151)(.8511)
P = 12 (.15)(0.1673432436896142578125)
P=0.3012178386413056640625

Here we can observe that the chances of finding exactly 1 producer well out of 12 wells drilled is approximately 30%.

thanks
 
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ORACLE said:
Questions
1. National Oil Company conducts exploratory oil drilling operations in the south western United States. To fund the operation, investors form partnerships, which provide the financial support necessary to drill a fixed number of oil wells. Each well drilled is classified as a producer well or a dry well. Past experience shows that 15% of all wells drilled are producer wells. A newly formed partnership has provided the financial support for drilling at 12 exploratory locations.

1.What is the probability that all 12 wells will be producer wells?

We have to find the probability of all the 12 wells drilled are producer wells with a 15% prior probability. Here we will calculate the probability using binomial distribution method.
The formula is
P(k out of n) =[n!/k!(n-k)! ](p^k)(q^n-k)
where k =12, number of times a producer well drilled, p = 15% or .15 is the observed probability of a producer well , q = 85% or .85 is the complementary probability (1-p) that is of a dry well, and n = 12 is the number of wells drilled.
Substituted as
P(12 12) =[479001600/479001600(1)](.1512)(.851)
P=0.000000000129746337890625
We can observe that the chances of finding all 12 wells as producer wells are approximately nill.
A couple of comments: your answer is numerically correct but you mean
(.15^12)(.85^0), not "(.1512)(.851) (and certainly not that last "1"!). Since your probabilties were only give to 2 decimal places, keeping 15 significant figures is misleading: P= 1.30 x 10^(-10) would be better.
(And I don't believe that "approximately nil" is a mathematical term!)

2.What is the probability that all 12 wells will be dry wells?

Here, we will find the probability of all the 12 wells drilled are dry wells with 85% prior probability.
k =12, number of times a dry well drilled, p = 85% or .85 is the observed probability of a dry well , q = 15% or .15 is the complementary probability (1-p) that is of a producer well, and n = 12 is the number of wells drilled.
Substituted as
P(12 out of 12) =[479001600/479001600(1)](.8512)(.151)
P=0.142241757136172119140625
We can observe a 14.2% chance of finding all 12 wells as dry wells.
Same comments as above: You mean (.85^12)(.15^0). I would have said "a 14% chance" since your original probabilities were given to the nearest percent.


3.What is the probability that exactly one well will be a producer well?

k =1, number of times a producer well drilled, p = 15% or .15 is the observed probability of a producer well , q = 85% or .85 is the complementary probability (1-p) that is of a dry well, and n = 12 is the number of wells drilled.
Substituted as
P(1/12) =[479001600/1(39916800)](.151)(.8511)
P = 12 (.15)(0.1673432436896142578125)
P=0.3012178386413056640625

Here we can observe that the chances of finding exactly 1 producer well out of 12 wells drilled is approximately 30%.[/B
thanks

Again, you mean (.15^1)(.85^11)- but now the "1" is correct!
 
hi HallsofIvy

missing carets and others were copy paste errors.

thanks for your help and suggestions
 
also Prof. HallsofIvy,

I have couple of sets of problems that i have solved which needs to be commented. Can I post those?
Iam in IT profession and doing my MBA in info systems.
thanks
 
You are welcome to post as long as you abide by forum rules: Make an effort to do them yourself and post your work.
 
probability questions & solutions

hi professor

questions and my answers are attached as pdf file.
pls comment.

thanks a lot
 

Attachments

All the problems look good except number three. I notice that when the problem says, for example, "At least 15", you are using "14.9". The standard "integer correction" is 1/2: any number larger than 14.5 would be rounded up to 15 so "at least 15" should be interpreted as 14.5.

By the way- it is not a good idea to append new questions to an old thread. Many people, such as myself, once they have read or responded to a thread do not look at it again.

It is also true that many people will not open files like the pdf file you attached for fear of viruses. It is better to take the trouble to type the information in.
 
thank u professor.
 

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