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

In summary: I'll keep it in mind. - I have not understood what "integer correction" is. can u pls explain me?- sure, I'll type them all from next time. thanks againInteger correction is the rule that when a decimal number is rounded to an integer, it is always rounded up if the decimal part is greater than or equal to 0.5 and down if it is less than 0.5. For example, 3.4 is closer to 3 than 4 so we round down to 3. 3.6 is closer to 4 than 3 so we round up to 4. But 3.5 is exactly half-way between 3 and 4 so we "round
  • #1
ORACLE
12
0
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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  • #2
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!
 
  • #3
hi HallsofIvy

missing carets and others were copy paste errors.

thanks for your help and suggestions
 
  • #4
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
 
  • #5
You are welcome to post as long as you abide by forum rules: Make an effort to do them yourself and post your work.
 
  • #6
probability questions & solutions

hi professor

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

thanks a lot
 

Attachments

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  • #7
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.
 
  • #8
thank u professor.
 

1. What is the difference between classical and quantum probability?

Classical probability deals with events that have a well-defined outcome, such as rolling a dice or flipping a coin. Quantum probability, on the other hand, deals with the probability of outcomes in quantum mechanical systems, which have inherent uncertainty due to the principles of quantum mechanics.

2. How is probability related to the principles of quantum mechanics?

Probability is a fundamental concept in quantum mechanics, as it is used to describe the likelihood of different outcomes in quantum systems. The principles of quantum mechanics, such as superposition and uncertainty, play a crucial role in determining the probabilities of different outcomes.

3. What is the role of probability in quantum computing?

Probability is a key component of quantum computing, as it is used to determine the likelihood of different qubit states and the success of quantum algorithms. The probabilistic nature of quantum systems allows for efficient computation and solving of complex problems.

4. How is quantum probability different from classical statistics?

Quantum probability differs from classical statistics in that it allows for the existence of superposition and entanglement, which are unique to quantum systems. These phenomena lead to the use of complex numbers and vector spaces in quantum probability, rather than the simpler probabilities and distributions used in classical statistics.

5. Can probability be used to predict the behavior of individual particles in quantum systems?

No, probability cannot be used to predict the behavior of individual particles in quantum systems. According to the principles of quantum mechanics, the exact state and behavior of a particle cannot be known, only the probability of different outcomes. Therefore, probability is used to describe the overall behavior of a system, rather than the behavior of individual particles.

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