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Chapter 5: Problem 42

A standard card deck has 52 cards consisting of 26 black and 26 red cards. Three cards are dealt from a shuffled deck, without replacement. a. True or false: The probability of being dealt three black cards is \((1 / 2) \times(1 / 2) \times(1 / 2)=1 / 8 .\) If true, explain why. If false, show how to get the correct probability. b. Let \(A=\) first card red and \(B=\) second card red. Are A and B independent? Explain why or why not. c. Answer parts a and b if each card is replaced in the deck after being dealt.

Short Answer

Expert verified
a. False, correct probability is \(\frac{12}{51}\). b. A and B are not independent. c. With replacement, probability is \(\frac{1}{8}\); A and B are independent.

Step by step solution

01

Understanding Part a

The exercise asks us if the probability calculation of being dealt three black cards, given as \((1/2) \times (1/2) \times (1/2) = 1/8\), is correct. The probability of drawing a black card changes as cards are drawn without replacement.
02

Calculating Correct Probability for Part a

The probability of drawing the first black card is \(\frac{26}{52} = \frac{1}{2}\). For the second black card, it becomes \(\frac{25}{51}\) and for the third black card \(\frac{24}{50}\). The correct probability is \(\frac{1}{2} \times \frac{25}{51} \times \frac{24}{50}\).
03

Solving Part a

Multiply the probabilities: \(\frac{1}{2} \times \frac{25}{51} \times \frac{24}{50} = \frac{300}{2550}\). Simplify to \(\frac{1}{17}\). Therefore, the statement is false.
04

Understanding Independence in Part b

Two events, A and B, are independent if the occurrence of one does not affect the probability of the other. We need to check if the probability of A and B occurring is the product of their probabilities separately.
05

Checking Independence for Part b

Probability of first card red, \(P(A) = \frac{26}{52} = \frac{1}{2}\). Probability of second card red given first is red, \(P(B|A) = \frac{25}{51}\). Probability of second card red, \(P(B) = \frac{26}{51}\). Since \(P(B|A) eq P(B)\), A and B are not independent.
06

Modifying for Replacement Scenario in Part c

With replacement, each draw is independent. Therefore, \(P(A) = \frac{26}{52} = \frac{1}{2}\) and \(P(B) = \frac{26}{52} = \frac{1}{2}\). Probability for part a is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\), and A and B become independent since \(P(B|A) = P(B)\).

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Key Concepts

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

Independent Events
In probability theory, independent events refer to situations where the occurrence of one event does not affect the probability of another event occurring.
When two events, say A and B, are independent, the probability of both occurring can be calculated by multiplying their individual probabilities:
  • For independent events: \( P(A \text{ and } B) = P(A) \times P(B) \)
If we consider drawing cards from a deck with replacement, each card draw is an independent event. The deck returns to its original state after each draw, thus the probability of drawing a specific card does not change. For example, if you draw a red card and then replace it, the probability of drawing another red card remains the same.
Probability without Replacement
Probability without replacement involves situations where the total number of outcomes changes after each event. This is because the objects (such as cards) are not returned to the pool after being chosen.
When dealing with a card deck, removing a card reduces the total number of available cards left for subsequent draws:
  • For example, the probability of drawing a black card initially is \( \frac{26}{52} \), but if the first card drawn is black, the probability of drawing another black card becomes \( \frac{25}{51} \).
Without replacement, each event is dependent on the previous events as the sample space changes. This concept is critical when calculating probabilities in succession, such as the probability of drawing three black cards without replacement which involves multiplying successive probabilities: \( \frac{26}{52} \times \frac{25}{51} \times \frac{24}{50} \).
Conditional Probability
Conditional probability refers to finding the probability of an event occurring given that another event has already occurred.
Mathematically, it is represented as \( P(B|A) \), which reads as 'the probability of B given A'.
This is calculated by dividing the probability of both events happening by the probability of the first event.
  • For instance, in a card deck, if you want to find the probability of drawing a red card (event B) after already drawing a red card (event A), it would be \( \frac{25}{51} \), assuming one red card has been drawn already.
Conditional probability shows us that the likelihood of certain outcomes can change based on preceding events, specifically when objects are not replaced.
Card Deck Probabilities
A standard card deck consists of 52 cards, split evenly into black and red suits.
Each suit consists of 13 cards, divided into four suits of clubs, diamonds, hearts, and spades.
  • When calculating probabilities with cards, it is crucial to consider whether the cards are drawn with or without replacement, as this affects the outcome matrix.
  • For example, if drawing without replacement, the probabilities continuously adjust since the cards in the deck decrease.
  • Conversely, when drawing with replacement, the probabilities for each draw remain the same as the card is put back into the deck. This maintains independence between events.
Understanding the structure of the card deck and the method of drawing is essential in evaluating the probability of different card outcomes accurately.

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Most popular questions from this chapter

A teacher announces a pop quiz for which the student is completely unprepared. The quiz consists of 100 true-false questions. The student has no choice but to guess the answer randomly for all 100 questions. a. Simulate taking this quiz by random guessing. Number a sheet of paper 1 to 100 to represent the 100 questions. Write a \(\mathrm{T}\) (true) or \(\mathrm{F}\) (false) for each question, by predicting what you think would happen if you repeatedly flipped a coin and let a tail represent a T guess and a head represent an F guess. (Don't actually flip a coin, but merely write down what you think a random series of guesses would look like.) b. How many questions would you expect to answer correctly simply by guessing? c. The table shows the 100 correct answers. The answers should be read across rows. How many questions did you answer correctly? d. The above answers were actually randomly generated by the Simulating the Probability of Head With a Fair Coin applet on the text CD. What percentage were true, and what percentage would you expect? Why are they not necessarily identical? e. Are there groups of answers within the sequence of 100 answers that appear nonrandom? For instance, what is the longest run of Ts or Fs? By comparison, which is the longest run of Ts or Fs within your sequence of 100 answers? (There is a tendency in guessing what randomness looks like to identify too few long runs in which the same outcome occurs several times in a row.)

Serena Williams won the 2010 Wimbledon Ladies' Singles Championship. For the seven matches she played in the tournament, her total number of first serves was \(379,\) total number of good first serves was 256 , and total number of double faults was 15 . a. Find the probability that her first serve is good. b. Find the conditional probability of double faulting, given that her first serve resulted in a fault. c. On what percentage of her service points does she double fault?

A single random digit is selected using software or a random number table. a. State the sample space for the possible outcomes. b. State the probability for each possible outcome, based on what you know about the way random numbers are generated. c. Each outcome in a sample space must have probability between 0 and 1 , and the total of the probabilities must equal 1. Show that your assignment of probabilities in part b satisfies this rule.

Each year, the Web site espn.go.com/nba/hollinger/playoffodds displays the probabilities of professional basketball teams achieving certain goals. For example, at the end of the \(2009-2010\) regular season, the site listed the following probabilities (expressed as percentages) of each of the 16 playoff teams winning the NBA Championship. (Note that the site uses the term odds to represent in this context a probability.) The Web site explains that a "computer plays out the remainder of the season 5000 times to see the potential range of projected outcomes." a. Note the sum of the probabilities for the 16 teams is 99.9. Why do you think the sum differs from \(100 ?\) b. Interpret Orlando's probability of \(29.5 \%,\) which was calculated from the 5000 simulations. Is it based on the relative frequency or the subjective interpretation of probability?

A local downtown arts and crafts shop found from past observation that \(20 \%\) of the people who enter the shop actually buy something. Three potential customers enter the shop. a. How many outcomes are possible for whether the clerk makes a sale to each customer? Construct a tree diagram to show the possible outcomes. \((\) Let \(Y=\) sale \(, N=\) no sale. \()\) b. Find the probability of at least one sale to the three customers. c. What did your calculations assume in part b? Describe a situation in which that assumption would be unrealistic.
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