/*! 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 15 Consider the experiment of selec... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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): \( P(Face \ Card) = \frac{12}{52} = \frac{3}{13} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

Consider the experiment of tossing a coin three times. a. Develop a tree diagram for the experiment. b. List the experimental outcomes. c. What is the probability for each experimental outcome?

High school seniors with strong academic records apply to the nation's most selective colleges in greater numbers each year. Because the number of slots remains relatively stable, some colleges reject more early applicants. Suppose that for a recent admissions class, an Ivy League college received 2851 applications for early admission. Of this group, it admitted 1033 students early, rejected 854 outright, and deferred 964 to the regular admission pool for further consideration. In the past, this school has admitted \(18 \%\) of the deferred early admission applicants during the regular admission process. Counting the students admitted early and the students admitted during the regular admission process, the total class size was \(2375 .\) Let \(E, R,\) and \(D\) represent the events that a student who applies for early admission is admitted early, rejected outright, or deferred to the regular admissions pool. a. Use the data to estimate \(P(E), P(R),\) and \(P(D)\). b. Are events \(E\) and \(D\) mutually exclusive? Find \(P(E \cap D)\). c. For the 2375 students who were admitted, what is the probability that a randomly selected student was accepted during early admission? d. Suppose a student applies for early admission. What is the probability that the student will be admitted for early admission or be deferred and later admitted during the regular admission process?

A local bank reviewed its credit card policy with the intention of recalling some of its credit cards. In the past approximately \(5 \%\) of cardholders defaulted, leaving the bank unable to collect the outstanding balance. Hence, management established a prior probability of .05 that any particular cardholder will default. The bank also found that the probability of missing a monthly payment is .20 for customers who do not default. Of course, the probability of missing a monthly payment for those who default is 1 a. Given that a customer missed one or more monthly payments, compute the posterior probability that the customer will default. b. The bank would like to recall its card if the probability that a customer will default is greater than \(.20 .\) Should the bank recall its card if the customer misses a monthly payment? Why or why not?

The Powerball lottery is played twice each week in 28 states, the Virgin Islands, and the District of Columbia. To play Powerball a participant must purchase a ticket and then select five numbers from the digits 1 through 55 and a Powerball number from the digits 1 through 42\. To determine the winning numbers for each game, lottery officials draw 5 white balls out of a drum with 55 white balls, and 1 red ball out of a drum with 42 red balls. To win the jackpot, a participant's numbers must match the numbers on the 5 white balls in any order and the number on the red Powerball. Eight coworkers at the ConAgra Foods plant in Lincoln, Nebraska, claimed the record \(\$ 365\) million jackpot on February \(18,2006,\) by matching the numbers \(15-17-43-44-49\) and the Powerball number \(29 .\) A variety of other cash prizes are awarded each time the game is played. For instance, a prize of \(\$ 200,000\) is paid if the participant's five numbers match the numbers on the 5 white balls (Powerball website, March 19,2006 ). a. Compute the number of ways the first five numbers can be selected. b. What is the probability of winning a prize of \(\$ 200,000\) by matching the numbers on the 5 white balls? c. What is the probability of winning the Powerball jackpot?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.