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Chapter 5: Problem 27

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

An article in the New York Times reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=0.01\) for people who suffer cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and difficulty finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The basis for the New York Times article was a research study of 2,329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2,329 cardiac arrest sufferers do you think survived? Explain.

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: \(P(\) Seed 1 defeats Seed 4\()=0.8\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) \(P(\) Seed 1 defeats \(\operatorname{Seed} 3)=0.7\) \(P(\) Seed 2 defeats \(\operatorname{Seed} 3)=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats Seed 4) \(=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

The following statement is from a letter to the editor that appeared in USA Today (September 3,2008 ): "Among Notre Dame's current undergraduates, our ethnic minority students \((21 \%)\) and international students \((3 \%)\) alone equal the percentage of students who are children of alumni (24\%). Add the \(43 \%\) of our students who receive need-based financial aid (one way to define working-class kids), and more than \(60 \%\) of our student body is composed of minorities and students from less affluent families." Do you think that the statement that more than \(60 \%\) of the student body is composed of minorities and students from less affluent families is likely to be correct? Explain.

There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\), and \(P(E \cap F)=0.15 .\) (Hint: See Example 5.5) a. Use the given probability information to set up a "hypothetical \(1000 "\) table with columns corresponding to \(E\) and not \(E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. the probability that Shelly must stop for at least one light (the probability of \(E \cup F)\). ii. the probability that Shelly does not have to stop at either light. iii. the probability that Shelly must stop at exactly one of the two lights. iv. the probability that Shelly must stop only at the first light.

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.
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