# An oil drilling company ventures into various locations, and

Brock Brown 2021-12-17 Answered
An oil drilling company ventures into various locations, and its success or failure is independent from one location to another. Suppose the probability of a success at any specific location is 0.30. What is the probability that a driller drills 10 locations and finds 1 success? The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller's prospects for bankruptcy? i. ii.
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## Expert Answer

Kirsten Davis
Answered 2021-12-18 Author has 27 answers
Step 1
Given Information:
Probability of success at any specific location i.e., $p=0.30$
i) To find the probability that a driller drills 10 locations and finds 1 success:
Let X be a random variable which denotes the number of successes among 10 drills. Because trials are independent, X follows a Binomial distribution with parameters .
Probability mass function of Binomial variable X is given as:
$P\left(X=x\right){=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}$
Required probability is obtained as follows:
$P\left(X=1\right){=}^{10}{C}_{1}{\left(0.30\right)}^{1}{\left(1-0.30\right)}^{10-1}$
$=\frac{10!}{1!\left(10-1\right)!}×0.30×{\left(0.70\right)}^{9}$
$=\frac{10×9!}{9!}×0.30×0.040353607$
$=10×0.0121060821$
$=0.121060821$
$\approx 0.1211$
Thus, the probability that a driller drills 10 locations and finds 1 success is 0.1211
Step 2
ii) Let X represent the number of drills required to get first success.
Find the probability that the driller prospects for bankruptcy.
In other words, that there are no successes $\left(X=0\right)$ in 10 trials:
Required probability is obtained as follows:
$P\left(X=0\right){=}^{10}{C}_{0}{\left(0.30\right)}^{0}{\left(1-0.30\right)}^{10-0}$
$=\frac{10!}{0!\left(10-0\right)!}×1×{\left(0.70\right)}^{10}$
$=\frac{10!}{10!}×1×0.0282475249$
$=0.0282475249$
$\approx 0.0282$
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godsrvnt0706
Answered 2021-12-19 Author has 31 answers
Suppose the probability of a success at any specific location is 0.25.
i) What is the probability that a driller drills 10 locations and finds 1 success?
$P\left(\text{one success in 10}\right)=\left(10C1\right){\left(0.25\right)}^{1}{\left(0.75\right)}^{9}=0.1877$...
ii) The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller is prospects for bankruptcy?
$P\left(\text{no success in 10 trials}\right)={0.75}^{10}=0.0563$...
Chances are 563 in 10000 that he will go bankrupt.
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nick1337
Answered 2021-12-28 Author has 510 answers

Step 1
- Consider that drilling in a location is a trial.
- Probability of success in each trial is p=0.25,
- Probability of failure is q=1-p=1-0.25=0.75
Step 2
a) Let random variable X represent the number of success among 10 drills.
Because trials are independent, X has binomial distribution with parameters n=10 and p=0.25.
Probability mass function of X is
$P\left(X=x\right)=Bin\left(x;10,0.25\right)$

Lets calculate the probability the driller drills at at 10 locations and has 1 success.
=0.1877
Step 3
b) Let random variable X represent the number of drills required to get first success.
Now, lets find the probability that the driller prospects for bankruptcy, in other words, that there are no successes in 10 trials.
$P\left(X=0\right)=\left(\begin{array}{c}10\\ 0\end{array}\right)\left(0.25{\right)}^{0}\left(0.75{\right)}^{10}$
=0.05631

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