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a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. \(\quad\) 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 \(B, P(A)=0.24, P(B)=0.24\) and \(P(A \cup B)=0.48\). Are \(A\) and \(B\) mutually exclusive? How can you tell?

Short Answer

Expert verified
i. The value of \(P(E \cap F)\) is 0. ii. The value of \(P(E \cup F)\) is 0.9. For part b, events A and B are mutually exclusive because their union's probability is equal to the sum of their individual probabilities.

Step by step solution

01

Determine if events are mutually exclusive

Given that events E and F are mutually exclusive, this means that they cannot both occur simultaneously. The probability of their intersection, \(P(E \cap F)\), should be 0.
02

Calculate the probability of the intersection of E and F

Since events E and F are mutually exclusive, the probability of their intersection is: \[P(E \cap F) = 0\]
03

Calculate the probability of the union of E and F

For mutually exclusive events, the probability of their union is the sum of their individual probabilities: \[P(E \cup F) = P(E) + P(F) = 0.14 + 0.76 = 0.9\] Now, let's analyze part b of the exercise.
04

Determine if events A and B are mutually exclusive

To check if events A and B are mutually exclusive, we need to compare the given probability of their union (\(P(A \cup B)\)) with the sum of their individual probabilities: \(P(A \cup B) = 0.48\) \(P(A) + P(B) = 0.24 + 0.24 = 0.48\) Since \(P(A \cup B) = P(A) + P(B)\), it indicates that events A and B are mutually exclusive.
05

Answer the questions

i. The value of \(P(E \cap F)\) is 0. ii. The value of \(P(E \cup F)\) is 0.9. For part b, events A and B are mutually exclusive because their union's probability is equal to the sum of their individual probabilities.

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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 the branch of mathematics that deals with the likelihood of events occurring. Probability is quantified as a number between 0 and 1, where 0 indicates an impossibility and 1 indicates certainty. Understanding the basics of probability is essential for analyzing different types of events, particularly when assessing outcomes in various endeavors like gambling, weather forecasting, and even in complex fields such as finance and the sciences.

When approaching problems in probability, it is crucial to define the experiment, identify possible outcomes, and determine the event of interest. This helps in assigning a probability value that measures the event's chance of happening. By mastering probability theory, students can better predict the likelihood of outcomes and make more informed decisions based on those predictions.
Intersection of Events
The intersection of events refers to a situation where two or more events happen at the same time. In mathematical terms, it's denoted by the symbol \(\cap\). In probability, the probability of the intersection of two events \(E\) and \(F\), written as \(P(E \cap F)\), helps us understand how likely it is for both events to occur together. If \(E\) and \(F\) are mutually exclusive, meaning they cannot both happen at the same time, then \(P(E \cap F) = 0\).

For example, in a deck of cards, the probability of drawing a card that is both red and a club (mutually exclusive events) is zero because a card cannot be from both suits simultaneously. Understanding intersections is crucial in calculating probabilities in more complex circumstances, such as those involving dependent events.
Union of Events
The union of two or more events, symbolized by \(\cup\), is the event that at least one of the included events occurs. When considering \(P(E \cup F)\), we're looking for the likelihood that event \(E\) or event \(F\), or both, happen. In the case of mutually exclusive events, which do not overlap, the probability of their union is simply the sum of their individual probabilities.

Calculating Union Probability

As in the original exercise, for mutually exclusive events \(E\) and \(F\), we get \(P(E \cup F) = P(E) + P(F)\). This concept is essential for determining the total probability of multiple outcomes and is frequently used across all areas of statistics and probability theory.
Mutually Exclusive
Mutually exclusive events are those that cannot occur simultaneously. If two events are mutually exclusive, the occurrence of one event excludes the possibility of the other event happening at the same time. This has significant implications in probability since it affects how probabilities are calculated.

In practical terms, if you're flipping a coin, the events 'Heads' and 'Tails' are mutually exclusive because the coin cannot land on both sides at the same time. As a result, mutually exclusive events have no elements in common, and their intersection is always an empty set, leading to \(P(E \cap F) = 0\). Recognizing mutually exclusive events is key to solving problems correctly and understanding the nature of how certain events relate to each other within a given context.

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Most popular questions from this chapter

An airline reports that for a particular flight operating daily between Phoenix and Atlanta, the probability of an on-time arrival is 0.86. Give a relative frequency interpretation of this probability.

In an article that appears on the website of the American Statistical Association, 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 $$ \text { i. } P(T D \mid D) $$ iii. \(P(C)\) $$ \text { ii. } P(T D \mid C) $$ 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 information in 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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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: $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=0.8 $$ \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 3)=0.7 $$ \(P(\) Seed 2 defeats Seed 3\()=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats \(\operatorname{Seed} 4)=0.6\) 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.

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