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Playing cards Shuffle a standard deck of playing cards and deal one card. Define events J: getting a jack, and R: getting a red card. (a) Construct a two-way table that describes the sample space in terms of events J and R. (b) Find \(P(I)\) and \(P(R)\) (c) Describe the event \(J\) and \(R\) in words. Then find \(P(J \text { and } R)\) (d) Explain why \(P(J \text { or } R) \neq P(J)+P(R) .\) Then use the general addition rule to compute \(P\left(\int \text { or } R\right)\)

Short Answer

Expert verified
The probability that a dealt card is either a jack or a red is \( \frac{7}{13} \).

Step by step solution

01

Define the Sample Space

A standard deck of playing cards consists of 52 cards. There are 4 jacks (one for each suit) and 26 red cards (13 hearts and 13 diamonds). The total number of possible outcomes when dealing one card is 52.
02

Construct the Two-Way Table

Create a table to display the intersection of events J (getting a jack) and R (getting a red card). - Jacks: 2 red jacks (hearts, diamonds) and 2 black jacks (spades, clubs). - Red Cards: 26 (13 hearts and 13 diamonds). | | Red Card (R) | Not Red | |-------|--------------|---------| | Jack (J) | 2 | 2 | | Not Jack | 24 | 24 |
03

Find Probabilities of Simple Events

To find the probability of getting a jack, calculate the number of jacks in the deck divided by the total number of cards, \( P(J) = \frac{4}{52} = \frac{1}{13} \).To find the probability of getting a red card, calculate the number of red cards in the deck divided by the total number of cards, \( P(R) = \frac{26}{52} = \frac{1}{2} \).
04

Describe and Find Probability for Joint Events

The event \( J \text{ and } R \) means getting a card that is both a jack and a red card. There are 2 red jacks, one of hearts and one of diamonds.So, \( P(J \text{ and } R) = \frac{2}{52} = \frac{1}{26} \).
05

Explanation and Use of Addition Rule

The event \( J \text{ or } R \) means getting a card that is either a jack or a red card. In probability, when two events can occur simultaneously and share common outcomes, the formula \( P(J \text{ or } R) = P(J) + P(R) - P(J \text{ and } R) \) is used. Applying the formula: \( P(J \text{ or } R) = \frac{1}{13} + \frac{1}{2} - \frac{1}{26} = \frac{2}{26} + \frac{13}{26} - \frac{1}{26} = \frac{14}{26} = \frac{7}{13} \).

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Key Concepts

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

Two-Way Table
A two-way table is a simple yet powerful tool used to organize data in probability and statistics, especially when dealing with two categorical variables. It helps in analyzing the relationship between these variables by arranging data into rows and columns.
A common example is the case of drawing cards from a deck, where one can categorize events based on different characteristics. In the current exercise, a two-way table is used to categorize whether a drawn card is a jack and whether it is red. The table has:
  • Rows representing whether a card is a jack or not.
  • Columns representing whether a card is red or not.
From this, you can easily see how each card fits into the categories. This makes it straightforward to find probabilities for combinations of these characteristics, like the probability of drawing a red jack. Understanding these tabular relationships is crucial for calculating probabilities accurately.
Addition Rule
The Addition Rule in probability helps compute the probability of either one event or another happening, especially when these events might have outcomes in common. This is different from simple addition, as common outcomes should not be counted twice.In our exercise, when calculating the probability of drawing either a jack or a red card or both, we must remember not to double-count the red jacks. Hence, the formula used is:
  • \[ P(J \text{ or } R) = P(J) + P(R) - P(J \text{ and } R) \]
Here:
  • \( P(J) \) is the probability of getting a jack.
  • \( P(R) \) is the probability of getting a red card.
  • \( P(J \text{ and } R) \) accounts for the overlap in our categories (red jacks).
Applying this rule ensures that the probability calculation is accurate and considers all possible outcomes efficiently.
Sample Space
The concept of sample space is a fundamental idea in probability theory. It represents the set of all possible outcomes of a particular experiment or random trial. Understanding the sample space provides the foundation for calculating any probabilities of individual events or combinations. In our card game scenario, the sample space consists of all 52 cards in a standard deck. This space can be further broken down into categories relevant to the problem, such as the four jacks (spades, hearts, diamonds, clubs) and the 26 red cards (hearts and diamonds).
The sample space allows us to calculate the likelihood of specific events, like drawing a red card or a jack, by examining how individual outcomes fit into these categories. By defining the sample space clearly, we can ensure that our probability calculations are based on an accurate assessment of all potential outcomes.
Joint Events
Joint events in probability refer to scenarios where two events occur simultaneously. They are crucial for understanding how different conditions can influence an outcome in an experiment or trial.Specifically, if one event is drawing a jack (J) and another is drawing a red card (R), the joint event \( J \text{ and } R \) involves drawing a card that fulfills both criteria — i.e., a card that is both a jack and red. This precise description narrows the set of possible outcomes to exactly those that occur at the intersection of these conditions.In our deck of playing cards, there are two such cards: the jack of hearts and the jack of diamonds. This joint probability can be calculated by determining the likelihood of this intersection: \( P(J \text{ and } R) \). Understanding joint events is essential for evaluating how combined conditions affect overall probabilities, aiding in a thorough comprehension of more complex probability situations.

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Most popular questions from this chapter

Tall people and basketball players Select an adult at random. Define events T: person is over 6 feet tall, and B: person is a professional basketball player. Rank the following probabilities from smallest to largest. Justify your answer. $$ P(T) \quad P(B) \qquad P(T | B) \qquad P(B | T) $$

Mac or PC? A recent census at a major university revealed that 40% of its students primarily used Macintosh computers (Macs). The rest mainly used PCs. At the time of the census, 67% of the school’s students were undergraduates. The rest were graduate students. In the census, 23% of respondents were graduate students who said that they used PCs as their main computers. Suppose we select a student at random from among those who were part of the census. (a) Assuming that there were 10,000 students in the census, make a two-way table for this chance process. (b) Construct a Venn diagram to represent this setting. (c) Consider the event that the randomly selected student is a graduate student who uses a Mac. Write this event in symbolic form using the two events of interest that you chose in (b). (d) Find the probability of the event described in (c). Explain your method.

Exercises 33 to 35 refer to the following setting. A basketball player makes 47% of her shots from the field during the season. You want to estimate the probability that the player makes 5 or more of 10 shots. You simulate 10 shots 25 times and get the following numbers of hits: 5754153434534463417455657 What is your estimate of the probability? (a) 5/25, or 0.20 (b) 11/25, or 0.44 (c) 12/25, or 0.48 (d) 16/25, or 0.64 (e) 19/25, or 0.76

Probability models? In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate, that is, satisfies the rules of probability. If not, give specific reasons for your answer. (a) Roll a die and record the count of spots on the up-face: \(P(1)=0, P(2)=1 / 6, P(3)=1 / 3, P(4)=1 / 3,\) \(P(5)=1 / 6, P(6)=0\) (b) Choose a college student at random and record gender and enrollment status: \(P(\text { female full-time })=\) \(0.56, P(\text { male full -time })=0.44, P(\text { female part-time })=\) \(0.24, P(\text { male part-time })=0.17\) (c) Deal a card from a shuffled deck: \(P(\text { clubs })=\) \(12 / 52, P\) (diamonds \()=12 / 52, P(\text { hearts })=12 / 52\) \(P(\text { spades })=16 / 52\) .

Multiple choice: Select the best answer for Exercises 104 to 106. An athlete suspected of using steroids is given two tests that operate independently of each other. Test A has probability 0.9 of being positive if steroids have been used. Test B has probability 0.8 of being positive if steroids have been used. What is the probability that neither test is positive if steroids have been used? $$ \begin{array}{ll}{\text { (a) } 0.72} & {\text { (c) } 0.02 \quad \text { (e) } 0.08} \\ {\text { (b) } 0.38} & {\text { (d) } 0.28}\end{array} $$

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