A geological study indicates that an exploratory oil well should strike oil with

Trent Carpenter

Trent Carpenter

Answered question

2021-09-15

A geological study indicates that an exploratory oil well should strike oil with probability of 0.2. What is the probability that
A. The first strike comes on the third well drilled?
B. The third strike comes on the seventh well drilled?
C. What assumptions did you make to obtain the answers to parts A and B?
D. Find the mean and variant of the number of wells that must be drilled if the company wants to set up three producing wells.

Answer & Explanation

Bentley Leach

Bentley Leach

Skilled2021-09-16Added 109 answers

Step 1
Given : A geological study indicates that an exploratory oil well should strike oil with probability 0.2.
Define random variable X denotes the number of oil wells drilled it takes to observe the third oil strike.
Striking probability : p=0.2
Since, each oil well has a constant 0.2 probability of being an oil strike. It is reasonable to assume that X has a negative binomial distribution.
X follows a negative binomial distribution with parameters r=3 (number of oil strikes needed) and p=0.2 (probability of observing an oil strike).
Notation: XNB(r=3,p=0.2)
The PMF is given as:
P(X=x)=nb(x;r,p)
={(x+r1 r1)pr(1p)xx=0,1,2,0otherwise
Step 2
(a) Given: p=0.2
q=1p=10.2=0.8
r=1
Using negative binomial probability formula,
p(y)=(beg{array}{c}y1r1end{array})prqyr
Evaluate:
p(3)=(beg{array}{c}3111end{array})0.210.831
=0.128
Probability that the first strike comes on the third well drilled is 0.128.
(b) Given: r=3,p=0.2,q=0.8
Using negative binomial probability formula,
p(y)=(beg{array}{c}y1r1end{array})prqyr
Evaluate:
p(3)=(beg{array}{c}7131end{array})0.230.873
=0.0492
Probability that the third strike comes on the seventh well drilled is 0.0492.
Step 3
(c) 1. The experiment consists of a sequence of independent trials.
2. Each trial can result in either a success (S) or a failure (F).
3. The probability of success is constant from trial to trial, so P(S on trial i) =p f or  i=1,2,
4. The experiment continues (trials are performed) until a total of r successes have been observed, where r is a specified positive integer.
(d) Given: r=3,p=0.2,q=0.8
The mean of a negative binomial distributed variable is r divided by p:
E(X)=rp=30.2=15
Mean =15
The variance of a negative binomial distributed variable is r(1p) divided by p2.
V(X)=r(1p)p2=3(10.2)0.22
V(X)=60
Variance =60

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