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Consider the experiment of selecting a playing card from a deck of 52 playing cards. Each card corresponds to a sample point with a \(1 / 52\) probability. a. List the sample points in the event an ace is selected. b. List the sample points in the event a club is selected. c. List the sample points in the event a face card (jack, queen, or king) is selected. d. Find the probabilities associated with each of the events in parts (a), (b), and (c).

Short Answer

Expert verified
Aces: Ace of Hearts, Diamonds, Clubs, Spades. Clubs: All 13 clubs. Face cards: 12 from all suits. Probabilities: \(\frac{1}{13}, \frac{1}{4}, \frac{3}{13}\).

Step by step solution

01

Understand the Deck

A standard deck of 52 playing cards contains 4 suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards: numbers 2 through 10, and the face cards Jack (J), Queen (Q), King (K), and the Ace (A).
02

List Aces

An ace can be selected from each of the four suits: hearts, diamonds, clubs, and spades. Therefore, the sample points for the event of selecting an ace are: Ace of Hearts, Ace of Diamonds, Ace of Clubs, and Ace of Spades.
03

List Clubs

The club suit contains 13 cards. So, the sample points for selecting a club are: 2 of Clubs, 3 of Clubs, 4 of Clubs ... 10 of Clubs, Jack of Clubs, Queen of Clubs, King of Clubs, and Ace of Clubs.
04

List Face Cards

Face cards in each suit are the Jack, Queen, and King. Since there are 4 suits, the sample points for selecting a face card are: Jack of Hearts, Queen of Hearts, King of Hearts; Jack of Diamonds, Queen of Diamonds, King of Diamonds; Jack of Clubs, Queen of Clubs, King of Clubs; Jack of Spades, Queen of Spades, King of Spades.
05

Find Probability of Selecting an Ace

The probability of selecting an ace is the number of aces (4) divided by the total number of cards (52): \( P(Ace) = \frac{4}{52} = \frac{1}{13} \).
06

Find Probability of Selecting a Club

The probability of selecting a club is the number of clubs (13) divided by the total number of cards (52): \( P(Club) = \frac{13}{52} = \frac{1}{4} \).
07

Find Probability of Selecting a Face Card

The probability of selecting a face card is the number of face cards (12) divided by the total number of cards (52): \( P(Face \ Card) = \frac{12}{52} = \frac{3}{13} \).

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Key Concepts

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

Sample Space
In the realm of probability theory, the sample space is a foundational concept. It encompasses all the possible outcomes of a random experiment. When dealing with a deck of 52 playing cards, our sample space comprises each individual card. This includes every suit and rank. Therefore, it involves all 13 hearts, 13 diamonds, 13 clubs, and 13 spades.
Each card represents a unique sample point within this space. In other words, every distinct card—be it the 2 of Hearts or the King of Spades—belongs to our set of potential outcomes.
Understanding this helps us realize that the deck's series of possible draws creates the backdrop against which probability functions. This complete set of sample points allows us to calculate probabilities for events, such as drawing an ace or a club.
Event
When we speak of an "event" in probability, we're referring to a specific subset of the sample space. It could be any defined condition or result that we focus on. For instance, selecting a card from the deck is a random experiment, and events could include specific scenarios like drawing an ace or a face card.
These events can include one or more sample points. For example, the event of drawing an ace consists of four sample points: Ace of Hearts, Ace of Diamonds, Ace of Clubs, and Ace of Spades. By defining events, we can narrow our focus on certain outcomes within the vast sample space, making probability calculations possible.
Events help translate the theoretical aspects of probability into real-world applications, enabling us to predict and analyze specific outcomes.
Random Experiment
A random experiment is a process that leads to one of several possible outcomes. The term "random" indicates that the outcome cannot be predicted with certainty. Each trial of the experiment can result in different outcomes based on chance.
Selecting a card from a deck exemplifies a common random experiment. With each draw, there's an element of uncertainty: what card will be picked? The setup is consistent—52 cards, 4 suits, each with 13 ranks—but the results vary with each card drawn.
Through repeated trials of this experiment, patterns and probabilities emerge, helping us predict likelihoods. This randomness is the core reason probability is both a fascinating and crucial field of study in mathematics and statistics.
Card Probability
Card probability explores chances associated with drawing certain cards from a deck. Understanding it requires knowledge of both the sample space and specific events. We calculate probabilities by comparing the number of favorable outcomes (events) to the total outcomes in the sample space.
For example, the probability of drawing an ace involves dividing the number of aces (4) by the total number of cards (52): \( P(Ace) = \frac{4}{52} = \frac{1}{13} \). This shows the likelihood of selecting an ace from a shuffled deck. Similarly, the probability of drawing a card from the club suit is \( P(Club) = \frac{13}{52} = \frac{1}{4} \).
Probabilities help quantify uncertainty, turning intuition into precise measures. Consequently, card probability is not just about chance but about accurately forecasting outcomes based on a defined mathematical framework.

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Most popular questions from this chapter

An experiment with three outcomes has been repeated 50 times, and it was learned that \(E_{1}\) occurred 20 times, \(E_{2}\) occurred 13 times, and \(E_{3}\) occurred 17 times. Assign probabilities to the outcomes. What method did you use?

The prior probabilities for events \(A_{1}\) and \(A_{2}\) are \(P\left(A_{1}\right)=.40\) and \(P\left(A_{2}\right)=.60 .\) It is also known that \(P\left(A_{1} \cap A_{2}\right)=0 .\) Suppose \(P\left(B | A_{1}\right)=.20\) and \(P\left(B | A_{2}\right)=.05\). a. Are \(A_{1}\) and \(A_{2}\) mutually exclusive? Explain. b. Compute \(P\left(A_{1} \cap B\right)\) and \(P\left(A_{2} \cap B\right)\). c. Compute \(P(B)\). d. Apply Bayes' theorem to compute \(P\left(A_{1} | B\right)\) and \(P\left(A_{2} | B\right)\).

Information about mutual funds provided by Morningstar Investment Research includes the type of mutual fund (Domestic Equity, International Equity, or Fixed Income) and the Morningstar rating for the fund. The rating is expressed from 1-star (lowest rating) to 5-star (highest rating). A sample of 25 mutual funds was selected from Morningstar Funds \(500(2008) .\) The following counts were obtained: \(\bullet\)Assume that one of these 25 mutual funds will be randomly selected in order to learn more about the mutual fund and its investment strategy. a. What is the probability of selecting a Domestic Equity fund? b. What is the probability of selecting a fund with a 4 -star or 5 -star rating? c. What is the probability of selecting a fund that is both a Domestic Equity fund and a fund with a 4 -star or 5 -star rating? d. What is the probability of selecting a fund that is a Domestic Equity fund or a fund with a 4 -star or 5 -star rating?Sixteen mutual funds were Domestic Equity funds. \(\bullet\)Thirteen mutual funds were rated 3 -star or less. \(\bullet\)Seven of the Domestic Equity funds were rated 4-star. \(\bullet\)Two of the Domestic Equity funds were rated 5 -star. Assume that one of these 25 mutual funds will be randomly selected in order to learn more about the mutual fund and its investment strategy. a. What is the probability of selecting a Domestic Equity fund? b. What is the probability of selecting a fund with a 4 -star or 5 -star rating? c. What is the probability of selecting a fund that is both a Domestic Equity fund and a fund with a 4 -star or 5 -star rating? d. What is the probability of selecting a fund that is a Domestic Equity fund or a fund with a 4 -star or 5 -star rating?

A financial manager made two new investments-one in the oil industry and one in municipal bonds. After a one-year period, each of the investments will be classified as either successful or unsuccessful. Consider the making of the two investments as an experiment. a. How many sample points exist for this experiment? b. Show a tree diagram and list the sample points. c. Let \(O=\) the event that the oil industry investment is successful and \(M=\) the event that the municipal bond investment is successful. List the sample points in \(O\) and in \(M\). d. List the sample points in the union of the events \((O \cup M)\). e. List the sample points in the intersection of the events \((O \cap M)\). f. Are events \(O\) and \(M\) mutually exclusive? Explain.

Statistics from the 2009 Major League Baseball season show that there were 157 players who had at least 500 plate appearances. For this group, 42 players had a batting average of 300 or higher, 53 players hit 25 or more home runs, and 14 players had a batting average of .300 or higher and hit 25 or more home runs. Only four players had 200 or more hits (ESPN website, January 10,2010 ). Use the 157 players who had at least 500 plate appearances to answer the following questions. a. What is the probability that a randomly selected player had a batting average of .300 or higher? b. What is the probability that a randomly selected player hit 25 or more home runs? c. Are the events having a batting average of .300 or higher and hitting 25 or more home runs mutually exclusive? d. What is the probability that a randomly selected player had a batting average of .300 or higher or hit 25 or more home runs? e. What is the probability that a randomly selected player had 200 or more hits? Does obtaining 200 or more hits appear to be more difficult than hitting 25 or more home runs? Explain.

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