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91Ó°ÊÓ

An oil exploration company currently has two active projects, one in Asia and the other in Europe. Let \(A\) be the event that the Asian project is successful and \(B\) be the event that the European project is successful. Suppose that \(A\) and \(B\) are independent events with \(P(A)=.4\) and \(P(B)=.7\). a. If the Asian project is not successful, what is the probability that the European project is also not successful? Explain your reasoning. b. What is the probability that at least one of the two projects will be successful? c. Given that at least one of the two projects is successful, what is the probability that only the Asian project is successful?

Short Answer

Expert verified
a. 0.18; b. 0.82; c. 0.1463

Step by step solution

01

Understanding the Problem

We have two projects, one in Asia (Event A) and one in Europe (Event B), which are independent. The probabilities given are \(P(A) = 0.4\) and \(P(B) = 0.7\). We'll find probabilities related to not being successful and at least one being successful, and then a conditional probability.
02

Calculating Probability of Both Projects Not Successful

To begin with, we calculate the probability of the Asian project not being successful, \(P(A')\), which is the complement of \(P(A)\). Use the formula \(P(A') = 1 - P(A)\), so \(P(A') = 1 - 0.4 = 0.6\). Similarly, find \(P(B') = 1 - P(B) = 0.3\). Since A and B are independent, \(P(A' \cap B') = P(A') \cdot P(B') = 0.6 \times 0.3 = 0.18\).
03

Calculating Probability of At Least One Project Successful

We use the formula for the probability of at least one event occurring: \(P(A \cup B) = 1 - P(A' \cap B')\). We previously found \(P(A' \cap B') = 0.18\), so \(P(A \cup B) = 1 - 0.18 = 0.82\).
04

Calculating Conditional Probability - Only Asian Project Successful

Now, we address finding the probability that only the Asian project is successful, given that at least one of the projects is successful. \(P(A \cap B')\) can be calculated as \(P(A)\cdot P(B') = 0.4 \times 0.3 = 0.12\). Using the conditional probability formula: \(P( ext{Only A} | A \cup B) = \frac{P(A \cap B')}{P(A \cup B)}\). Plugging in the values gives \(\frac{0.12}{0.82} \approx 0.1463\).

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Key Concepts

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

Probability Theory
Probability Theory is a branch of mathematics that deals with the likelihood of different outcomes in uncertain situations. At its core, it uses the concept of a sample space, which encompasses all possible outcomes of a given scenario. Each outcome in the sample space can be assigned a probability, which is a number between 0 and 1, indicating how likely it is to occur.
By using probability, we can gain insights into the likelihood of various events, allowing for better decision making under uncertainty. In the context of the exercise, the company predicts the success of oil exploration projects using probability calculations.
This helps them to understand the risk and potential return of their investments. Basic probability operations involve events like union, intersection, and complements, which help in making such forecasts.
Complementary Events
Complementary events are pairs of events in which one event happens if and only if the other does not happen. These are mutually exclusive and exhaustive events. In probability equations, if you have the probability of an event, you can easily find the probability of its complement.
  • The formula to calculate the complement of an event is: \(P(A') = 1 - P(A)\).
  • Here, \(A'\) is the event that \(A\) does not happen.
In the exercise, for instance, if the Asian project has a 40% chance of success, it means there's a 60% chance it will not succeed. Understanding complementary events allows businesses to prepare for unfavorable outcomes effectively.
Conditional Probability
Conditional Probability concerns the probability of an event occurring, given that another event has already happened. It's crucial in scenarios where events might be dependent or when considering a sequence of events.
The formula that expresses conditional probability is:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
This reads as the probability of event A occurring given that B has occurred. In our exercise, the company might be interested in the success of the Asian project given information about success in general.
Determining conditional probabilities helps refine predictions by factoring in new, relevant information.
Probability of Union of Events
The Probability of Union of Events deals with the chance of at least one of multiple events occurring. Symbolically, this is represented as \(P(A \cup B)\), which denotes the probability of event A or event B, or both happening.
The formula for calculating this probability is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
In scenarios where the events are independent, as in our exercise, the formula simplifies by using the complements: \(P(A \cup B) = 1 - P(A' \cap B')\). For the company, knowing this probability helps in understanding the likelihood of success with either project.
This insight can guide decisions on how resources should be allocated to maximize the chances of at least one project's success.

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Most popular questions from this chapter

Ann and Bev have each applied for several jobs at a local university. Let \(A\) be the event that Ann is hired and let \(B\) be the event that Bev is hired. Express in terms of \(A\) and \(B\) the events a. Ann is hired but not Bev. b. At least one of them is hired. c. Exactly one of them is hired.

Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers 1 , \(2, \ldots, 15\) to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced? c. What is the probability that two phones of each type are among the first six serviced?

Prove that if \(P(B \mid A)>P(B)\) [in which case we say that "A attracts \(B\) "], then \(P(A \mid B)>P(A)\) [" \(B\) attracts \(A\) "].

Two pumps connected in parallel fail independently of each other on any given day. The probability that only the older pump will fail is 10 , and the probability that only the newer pump will fail is \(.05\). What is the probability that the pumping system will fail on any given day (which happens if both pumps fail)?

An individual is presented with three different glasses of cola, labeled \(C, D\), and \(P .\) He is asked to taste all three and then list them in order of preference. Suppose the same cola has actually been put into all three glasses. a. What are the simple events in this ranking experiment, and what probability would you assign to each one? b. What is the probability that \(C\) is ranked first? c. What is the probability that \(C\) is ranked first and \(D\) is ranked last?

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