/*! 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 27 a. Suppose events \(E\) and \(F\... [FREE SOLUTION] | 91Ó°ÊÓ

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a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.41\) and \(P(E)=0.23\). i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F) ?\) b. Suppose that for events \(A\) and \(B, P(A)=0.26, P(B)=0.34\), and \(P(A \cup B)=0.47 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Short Answer

Expert verified
i. \(P(E \cap F) = 0\), ii. \(P(E \cup F) = 0.64\); Events \(A\) and \(B\) are not mutually exclusive

Step by step solution

01

Find \(P(E \cap F)\) for mutually exclusive events

From the definition of mutually exclusive events, we know that two events cannot occur simultaneously. Thus, \(P(E \cap F) = 0\)
02

Find \(P(E \cup F)\) for mutually exclusive events

The union of two mutually exclusive events is found by adding their individual probabilities. Thus, \(P(E \cup F) = P(E) + P(F) = 0.41 + 0.23 = 0.64\)
03

Determine if events \(A\) and \(B\) are mutually exclusive

We know that if events are mutually exclusive then \(P(A \cup B) = P(A) + P(B)\). Let's compare \(0.26 + 0.34\) with \(0.47\). Since \(0.26 + 0.34 = 0.6\), which is not equal to \(0.47\), therefore events \(A\) and \(B\) are not mutually exclusive.

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Key Concepts

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

Probability of Union of Events
Understanding the probability of the union of events involves evaluating the likelihood of either one event, another event, or both events occurring at the same time. Mathematically, for any two events A and B, this is represented as \(P(A \cup B)\), which reads as the probability of A or B occurring.

Now, to calculate \(P(A \cup B)\), we generally add the probabilities of A and B and subtract the probability of their intersection, which is \(P(A \cap B)\). This subtraction is necessary because if we simply added the probabilities of A and B without considering the intersection, we would be counting the intersection twice. However, there's an exception to this rule when dealing with mutually exclusive events, which we'll explore in other sections.
Probability of Intersection of Events
The probability of the intersection of events refers to the likelihood of two or more events occurring simultaneously. Represented as \(P(A \cap B)\), it illustrates the intersection of event A with event B.

However, to determine this intersection, we need to consider if the events can actually occur at the same time. For independent events, where one event's occurrence doesn't affect the other's, we can find the intersection by multiplying the probabilities of the individual events (i.e., \(P(A) \times P(B)\)). But the scenario is different for mutually exclusive events; their intersection probability is always zero because they cannot occur simultaneously.
Definition of Mutually Exclusive Events
Mutually exclusive events are cases in which the occurrence of one event completely prevents the occurrence of another. In other words, these events cannot happen at the same time.

For example, when you flip a coin, the event of landing tails is mutually exclusive from the event of landing heads. As a result, if two events E and F are mutually exclusive, then by definition, \(P(E \cap F) = 0\), meaning the probability of their intersection is zero. This also simplifies the calculation of their union, since \(P(E \cup F) = P(E) + P(F)\), with no need to subtract anything for their intersection because it's non-existent.
Probability Principles
There are core principles in probability that provide the foundation for understanding the behavior of random phenomena. One fundamental principle is the addition rule, which helps in determining the probability of the union of two events, considering if they are mutually exclusive or not.

The multiplication rule is another key principle, used to calculate the probability of the intersection of independent events. Moreover, it's important to distinguish between independent and mutually exclusive events, as this affects how we apply these principles: for mutually exclusive events, there is no overlap, while independent events can happen at the same time without influencing each other. Understanding and correctly applying these principles is essential for accurate probability calculations.

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Most popular questions from this chapter

If you were to roll a fair die 1,000 times, about how many sixes do you think you would observe? What is the probability of observing a six when a fair die is rolled?

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In an article that appears on the website of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in \(100 .\) And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 - are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty \(T C=\) event that the test result is clean \(D=\) event that the person tested is actually dirty \(C=\) event that the person tested is actually clean a. Using the information in the quote, what are the values of i. \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the probabilities from Part (a) to construct a "hypothetical 1000 " table. c. What is the value of \(P(T D)\) ? d. Use the table to calculate the probability that a person is clean given that the test result is dirty, \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

A professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E} .\) One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?

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