/*! 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 113 A single card is drawn from a st... [FREE SOLUTION] | 91Ó°ÊÓ

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A single card is drawn from a standard deck. Let A be the event that "the card is a face card" (a jack, a queen, or a king), \(\mathbf{B}\) is a "red card," and \(\mathrm{C}\) is "the card is a heart." Determine whether the following pairs of events are independent or dependent: a. \(\quad A\) and \(B\) b. \(\mathrm{A}\) and \(\mathrm{C}\) c. \(\quad B\) and \(C\)

Short Answer

Expert verified
The pairs of events \(A\) and \(B\), \(A\) and \(C\) are independent, while the pair \(B\) and \(C\) is dependent.

Step by step solution

01

Define The Events

In this problem, there are three events. Event A is getting a face card which are jacks, queens, and kings. There are 12 face cards in a standard 52-card deck. Event B is drawing a red card. Since half the deck is hearts or diamonds which are the red suits, there are 26 red cards in the deck. Event C is drawing a heart which are the 13 cards of the hearts suite.
02

Analyze Independence of Event A and B

The probability of event A, getting a face card, is \(\frac{12}{52} = \frac{3}{13}\). The probability event B, getting a red card, is \(\frac{26}{52} = \frac{1}{2}\). The probability of both events A and B happening, getting a red face card, is \(\frac{6}{52} = \frac{3}{26}\). These events would be independent if \(P(A)P(B)=P(A \cap B)\). However, \(\frac{3}{13} * \frac{1}{2} = \frac{3}{26}\) equals to the probability of both events happening, so events A and B are independent.
03

Analyze Independence of Event A and C

The probability of event A is \(\frac{3}{13}\). The probability of event C, getting a heart, is \(\frac{13}{52} = \frac{1}{4}\). The probability of both events A and C happening, getting a face card that is also a heart, is \(\frac{3}{52}\). Checking if \(P(A)P(C)=P(A \cap C)\), we find \(\frac{3}{13} * \frac{1}{4} = \frac{3}{52}\). Events A and C are thus independent.
04

Analyze Independence of Event B and C

The probability of event B is \(\frac{1}{2}\). The probability of event C is \(\frac{1}{4}\). The probability of both events B and C happening, getting a red heart, is \(\frac{13}{52} = \frac{1}{4}\). Checking if \(P(B)P(C)=P(B \cap C)\), we find \(\frac{1}{2} * \frac{1}{4} = \frac{1}{8}\), which does not equal the probability of both events happening, therefore events B and C are not independent but dependent.

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Key Concepts

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

Event Independence
Understanding the concept of event independence is key to mastering many problems in probability theory. When we talk about independent events in a probabilistic context, we are referring to the situation where the occurrence of one event does not influence the occurrence of another.

For example, consider the act of flipping a coin. The result of one flip does not impact the result of the next flip - each flip is independent. In our exercise, we determine independence by checking if the probability of event A occurring multiplied by the probability of event B occurring equals the probability of both events A and B occurring simultaneously. If this statement holds true, we confirm that events A and B are indeed independent.

As shown in the solution, by calculating these probabilities and applying the multiplication rule, we find that events A (getting a face card) and B (getting a red card) are independent. Also, events A and C (getting a heart) are independent. However, B and C are not because the combined probability does not match the product of their individual probabilities. Gaining a strong understanding of event independence is crucial for students as it paves the way to solving more complex probability problems efficiently.
Probability Theory
Probability theory is a mathematical framework for quantifying the uncertainty of various events. It is an essential part of statistics and underpins a wide range of disciplines, including finance, science, and engineering. In the context of our exercise, the probability theory helps us understand and calculate the likelihood of drawing certain cards from a standard deck.

The basics of probability can be summarized by saying that the probability of an event is a measure of the likelihood that the event will occur, expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the example of the card game, calculating the probability of drawing a face card (event A) or a red card (event B) requires an understanding of how many face and red cards are in the deck, allowing us to express their probabilities as \( \frac{12}{52} \) and \( \frac{26}{52} \) respectively.

Probability theory is not just about single events. It also involves understanding joint probabilities, conditional probabilities, and the independence of events, which are key aspects when determining the relationships between different events. The probability theory principles enable us to conclude with the solution involving the formulas that determine whether the given card events are independent or dependent.
Combinatorics
Combinatorics is a branch of mathematics concerning the study of countable, discrete structures and the counting, arrangement, and combination of sets of elements. It is closely linked with probability theory since many probability problems involve combinatorial calculations to determine the size of sample spaces and event spaces.

In our card-drawing scenario, combinatorics comes to play when we analyze the various possible outcomes. For instance, the total number of face cards, red cards, and hearts in a standard deck are all determined using combinatorial logic. A standard deck has 52 cards, which sets the foundation for our combinatorial calculations. For example, to obtain the probability of drawing a face card, we count the number of ways to draw one face card (12) out of the total number of cards in the deck (52), which gives us a probability of \( \frac{12}{52} \) or \( \frac{3}{13} \).

Understanding combinatorics enhances one's ability to solve problems because it involves analyzing how different events can be grouped, paired, or combined, and thus it pertains directly to the concept of event independence. Students benefit from learning combinatorial tactics as they frequently apply these to analyze potential outcomes in games of chance, computer science algorithms, and even in everyday decision-making processes.

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Most popular questions from this chapter

\(\mathrm{A}\) coin is flipped three times. a. Draw a tree diagram that represents all possible outcomes. b. Identify all branches that represent the event "exactly one head occurred." c. Find the probability of "exactly one head occurred."

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A woman and a man (unrelated) each has two children. At least one of the woman's children is a boy, and the man's older child is a boy. Is the probability that the woman has two boys greater than, equal to, or less than the probability that the man has two boys? a. Demonstrate the truth of your answer by using a simple sample to represent each family. b. Demonstrate the truth of your answer by taking two samples, one from men with two-children families and one from women with two-children families. c. Demonstrate the truth of your answer using computer simulation. Using the Bernoulli probability function with \(p=0.5 \text { (let } 0=\text { girl and } 1=\text { boy })\), generate 500 "families of two children" for the man and the woman. Determine which of the 500 satisfy the condition for each and determine the observed proportion with two boys. d. Demonstrate the truth of your answer by repeating the computer simulation several times. Repeat the simulation in part c several times. e. Do the preceding procedures scem to yicld the same results? Explain.

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