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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

5.62 An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. In Exercise \(5.34,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. b. The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer.

A large cable company reports the following: \(80 \%\) of its customers subscribe to cable TV service \(42 \%\) of its customers subscribe to Internet service \(32 \%\) of its customers subscribe to telephone service \(25 \%\) of its customers subscribe to both cable TV and Internet service \(21 \%\) of its customers subscribe to both cable TV and phone service \(23 \%\) of its customers subscribe to both Internet and phone service \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. In Exercise \(5.53,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. \(P(\) cable TV only) b. \(P\) (Internet \(\mid\) cable TV) c. \(P(\) exactly two services \()\) d. \(P\) (Internet and cable TV only)

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=0.4\) and \(P\left(E_{2}\right)=0.3\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\operatorname{not} E_{1}\) and not \(E_{2}\) ). c. What is the probability that the firm is successful in at least one of the two bids?

A study of the impact of seeking a second opinion about a medical condition is described in the paper "Evaluation of Outcomes from a National Patient- Initiated Second-Opinion Program". Based on a review of 6791 patient-initiated second opinions, the paper states the following: "Second opinions often resulted in changes in diagnosis (14.8\%), treatment \((37.4 \%),\) or changes in both \((10.6 \%)\)." Consider the following two events: \(D=\) event that second opinion results in a change in diagnosis \(T=\) event that second opinion results in a change in treatment a. What are the values of \(P(D), P(T),\) and \(P(D \cap T) ?\) b. Use the given probability information to set up a hypothetical 1000 table with columns corresponding to \(D\) and \(D^{C}\) and rows corresponding to \(T\) and \(T^{C}\). c. What is the probability that a second opinion results in neither a change in diagnosis nor a change in treatment? d. What is the probability that a second opinion results is a change in diagnosis or a change in treatment?

What does it mean to say that the probability that a coin toss will land head side up is \(0.5 ?\)

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