/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 Past experience has shown that, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Past experience has shown that, on the average, only 1 in 10 wells drilled hits oil. Let \(x\) be the number of drillings until the first success (oil is struck). Assume that the drillings represent independent events. a. Find \(p(1), p(2)\), and \(p(3)\). b. Give a formula for \(p(x)\). c. Graph \(p(x)\).

Short Answer

Expert verified
The probability of striking oil for the first time on the third well drilled can be found using the geometric probability distribution formula. The probability in this case is given by: $$ p(3) = (1-p)^{(3-1)}p = (0.9)^{(3-1)}(0.1) = 0.9^2(0.1) = 0.081 $$ Hence, the probability of striking oil for the first time on the third well drilled is 0.081 or 8.1%.

Step by step solution

01

Understand the geometric probability distribution formula

The geometric probability distribution is a discrete probability distribution that describes the number of Bernoulli trials needed to get the first success (\(x\)) in a sequence of independent Bernoulli trials, each with the same probability of success (\(p\)). The formula for the geometric probability distribution is: $$ p(x) = (1-p)^{(x-1)}p $$ where \(x\) is the number of trials until the first success, and \(p\) is the probability of success in each trial. In this exercise, the probability of hitting oil (success) is \(1/10\) or \(0.1\).
02

Calculate p(1), p(2), and p(3) using the geometric probability distribution formula

We have \(p = 0.1\). Now we will calculate the probabilities \(p(1)\), \(p(2)\), and \(p(3)\) using the formula: a) \(p(1) = (1-0.1)^{(1-1)}(0.1) = 0.1\) b) \(p(2) = (1-0.1)^{(2-1)}(0.1) = 0.9(0.1) = 0.09\) c) \(p(3) = (1-0.1)^{(3-1)}(0.1) = 0.9^2(0.1) = 0.081\)
03

Give a formula for p(x)

The formula for \(p(x)\) is based on the geometric probability distribution formula: $$ p(x) = (1-p)^{(x-1)}p $$ For this problem, we have \(p = 0.1\). So the formula for \(p(x)\) is: $$ p(x) = (0.9)^{(x-1)}(0.1) $$
04

Graph p(x)

To graph the function \(p(x)\), we can create a table of values of \(x\) and \(p(x)\). We can represent the discrete function using horizontal lines that represent the probability of each value of \(x\). The graph will be right-skewed because the probability of success is low (only \(10\%\)), and it decreases as the number of trials increases. Here is a simple table of values for visualization: | \(x\) | \(p(x)\) | |-----|-------| | 1 | 0.1 | | 2 | 0.09 | | 3 | 0.081 | | 4 | 0.073 | | ... | ... | Now we can use this table of values to graph the function \(p(x)\). The graph will consist of horizontal lines at each value of \(x\), with the length of the line proportional to the value of \(p(x)\). The line lengths decrease as \(x\) increases, indicating that the probability of striking oil decreases as more wells are drilled without success.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discrete Probability Distribution
In a discrete probability distribution, we deal with scenarios involving distinct outcomes. Unlike continuous probability distributions that can take any value within a range, discrete distributions are concerned with specific and countable events. Here, the geometric probability distribution is a type of discrete distribution.
This distribution is used when we are identifying the number of Bernoulli trials needed to achieve the first success. These outcomes, such as hitting oil or not per drilling in our exercise, are countable and distinct events. Thus, when you're asked about discrete probability distributions, think about setups where outcomes can be neatly listed and counted.
Bernoulli Trials
Bernoulli trials refer to a series of experiments that have precisely two outcomes: success or failure. Each trial is identical, meaning they follow the same rules and conditions, and crucially, the probability of success remains constant. These trials are named after Jacob Bernoulli, a prominent mathematician.
In the context of our problem, each attempt to drill for oil is a Bernoulli trial. You either find oil (success) or you don't (failure). The exercise assumes the probability of finding oil is 0.1, consistently across trials. Bernoulli trials are foundational to understanding various probability distributions, including the geometric distribution.
Probability of Success
Understanding the probability of success is central to solving problems involving Bernoulli trials. It's the likelihood that each individual trial results in success, labeled as \( p \).
For the exercise, the probability of hitting oil when drilling a well is \( 0.1 \) or 10%. This probability doesn't change from trial to trial, reflecting the constant chance of success, a crucial feature of Bernoulli trials.
The geometric probability distribution formula \( p(x) = (1-p)^{(x-1)}p \) uses this probability of success. In the formula, \( x \) denotes how many trials it takes to achieve the first success, and \( p \) is consistently 0.1 in our example. This constancy ensures the calculation of probabilities remains straightforward as we work through each trial.
Independent Events
A fundamental concept in probability theory is that of independent events. Two events are independent if the occurrence of one does not affect the probability of the other occurring. This means each event stands alone without influence from previous events.
In the context of drilling wells, each attempt is an independent event. The outcome of one drilling does not impact the next one. Whether or not you hit oil in the first well, the chance of finding oil in the next well remains the same, at 0.1 or 10% in our exercise.
This independence is critical when applying the geometric probability distribution model, as it ensures that calculations for each trial remain valid and unaffected by past outcomes. Understanding this independence helps in correctly applying probability formulas and accurately predicting outcomes in similar scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Permutations Evaluate the following permutations. a. \(P_{3}^{5}\) b. \(P_{9}^{10}\) c. \(P_{6}^{6}\) d. \(P_{1}^{20}\)

A population can be divided into two subgroups that occur with probabilities \(60 \%\) and \(40 \%,\) respectively. An event \(A\) occurs \(30 \%\) of the time in the first subgroup and \(50 \%\) of the time in the second subgroup. What is the unconditional probability of the event \(A\), regardless of which subgroup it comes from?

Suppose that, in a particular city, airport \(A\) handles \(50 \%\) of all airline traffic, and airports \(B\) and \(C\) handle \(30 \%\) and \(20 \%,\) respectively. The detection rates for weapons at the three airports are \(.9, .8,\) and \(.85,\) respectively. If a passenger at one of the airports is found to be carrying a weapon through the boarding gate, what is the probability that the passenger is using airport \(A\) ? Airport \(C ?\)

A piece of electronic equipment contains six computer chips, two of which are defective. Three chips are selected at random, removed from the piece of equipment, and inspected. Let \(x\) equal the number of defectives observed, where \(x=0,1,\) or 2 . Find the probability distribution for \(x\). Express the results graphically as a probability histogram.

A survey of people in a given region showed that \(20 \%\) were smokers. The probability of death due to lung cancer, given that a person smoked, was roughly 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is \(.006,\) what is the probability of death due to lung cancer given that a person is a smoker?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.