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Chapter 4: Problem 15

You draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. (a) Are the outcomes on the two cards independent? Why? (b) Find \(P(\) ace on 1 st card and king on 2 nd). (c) Find \(P(\) king on 1 st card and ace on 2 nd ). (d) Find the probability of drawing an ace and a king in either order.

Short Answer

Expert verified
Yes, the outcomes are independent. Probabilities: (a) independent, (b) \(\frac{1}{169}\), (c) \(\frac{1}{169}\), (d) \(\frac{2}{169}\).

Step by step solution

01

Understanding Independence of Events

Two events are independent if the outcome of one event does not affect the outcome of the other. In this exercise, because the first card is returned and the deck is reshuffled before drawing the second card, the outcomes on the two cards are independent.
02

Calculating the Probability of Drawing an Ace First and a King Second

The probability of drawing an ace from a standard deck of 52 cards is \( \frac{4}{52} \) because there are 4 aces in the deck. Once the card is replaced and the deck is reshuffled, the probability of drawing a king is \( \frac{4}{52} \). The events are independent, so we multiply these probabilities: \( P(\text{ace on 1st card and king on 2nd}) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169} \).
03

Calculating the Probability of Drawing a King First and an Ace Second

Similar to the previous step, the probability of drawing a king first is \( \frac{4}{52} \) and, after reshuffling, the probability of drawing an ace is \( \frac{4}{52} \). Hence, \( P(\text{king on 1st card and ace on 2nd}) = \frac{4}{52} \times \frac{4}{52} = \frac{1}{169} \).
04

Calculating Total Probability of Drawing an Ace and a King in Either Order

To find the probability of drawing an ace and a king in either order, we add the probabilities from Steps 2 and 3. Thus, \( P(\text{ace and king in either order}) = \frac{1}{169} + \frac{1}{169} = \frac{2}{169} \).

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Key Concepts

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

Independent Events in Probability
In probability theory, two events are considered independent if the occurrence of one does not affect the probability of the other occurring. For example, if you draw a card from a deck, replace it, and then draw another card, the result of the first draw does not influence the second. This is because the deck is reshuffled back to its original state.

When two events are independent:
  • The probability of both occurring is the product of their individual probabilities. For instance, if event A has a 1 in 5 chance of occurring, and event B has a 1 in 10 chance, the probability of both A and B occurring is (1/5) 脳 (1/10).
  • This principle is fundamental in understanding many probabilistic experiments, like the card draw scenario described.
The independence of events is what makes probabilistic calculations straightforward and manageable. When counting scenarios where independence does not hold, more complex methods like conditional probability are required.
Understanding Conditional Probability
Conditional probability is a measure of the probability of an event occurring, given that another event has already occurred. This concept is crucial when dealing with dependent events, where the outcome of one affects the likelihood of the other.

With the card draw exercise discussed, conditional probability isn't necessary since each draw is independent due to reshuffling. However, consider if you didn't replace the card:
  • The probability of drawing an ace then getting a king changes. Once you have an ace, less non-aces remain, affecting subsequent draws.
  • The formula for conditional probability becomes handy: \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \), where \( P(A \mid B) \) is the probability of A given B.
It's helpful for understanding dependencies in experiments and everyday scenarios.
Card Probability Basics
Probabilities involving cards are a common way to illustrate fundamental probability concepts. A standard deck of cards contains 52 cards, divided equally among four suits: spades, hearts, diamonds, and clubs. Each suit has 13 ranks, from Ace through King.

To find the probability of drawing a specific card like an Ace:
  • There are 4 Aces in the deck and 52 total cards.
  • Thus, the probability of drawing an Ace is \( \frac{4}{52} \), which simplifies to \( \frac{1}{13} \).
  • Similarly, the probability for any specific card (e.g., a King) follows the same logic.
This common scenario is often used to explain more complex probability topics such as combinations and permutations.
Understanding a Deck of Cards
A standard deck of cards, commonly used in a myriad of games, provides a rich framework for exploring probability. Each deck has:
  • 52 cards in total.
  • 4 suits: spades, hearts (red), and diamonds, clubs (black).
  • Each suit includes 13 ranks: Ace through to King.
  • Specific cards like Jacks, Queens, and Kings are known as face cards.
Knowing the structure of a deck is crucial for solving probability problems related to card games. Understanding how cards are organized allows one to apply probability rules intuitively and accurately. This organization also supports various strategies in card games, providing a mathematical basis for predicting and planning possible moves or outcomes.

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Most popular questions from this chapter

For each of the following situations, explain why the combinations rule or the permutations rule should be used. (a) Determine the number of different groups of 5 items that can be selected from 12 distinct items. (b) Determine the number of different arrangements of 5 items that can be selected from 12 distinct items.

Are customers more loyal in the East or in the West? The following table is based on information from Trends in the United States, published by the Food Marketing Institute, Washington, D.C. The columns represent length of customer loyalty (in years) at a primary supermarket. The rows represent regions of the United States. $$\begin{array}{lcrrrrrr} \hline & \begin{array}{c} \text { Less Than } \\ 1 \text { Year } \end{array} & \begin{array}{c} 1-2 \\ \text { Years } \end{array} & \begin{array}{c} 3-4 \\ \text { Years } \end{array} & \begin{array}{c} 5-9 \\ \text { Years } \end{array} & \begin{array}{c} 10-14 \\ \text { Years } \end{array} & \begin{array}{c} 15 \text { or More } \\ \text { Years } \end{array} & \begin{array}{l} \text { Row } \\ \text { Total } \end{array} \\ \hline \text { East } & 32 & 54 & 59 & 112 & 77 & 118 & 452 \\ \text { Midwest } & 31 & 68 & 68 & 120 & 63 & 173 & 523 \\ \text { South } & 53 & 92 & 93 & 158 & 106 & 158 & 660 \\ \text { West } & 41 & 56 & 67 & 78 & 45 & 86 & 373 \\ \text { Column Total } & 157 & 270 & 287 & 468 & 291 & 535 & 2008 \\ \hline \end{array}$$ What is the probability that a customer chosen at random (a) has been loyal 10 to 14 years? (b) has been loyal 10 to 14 years, given that he or she is from the East? (c) has been loyal at least 10 years? (d) has been loyal at least 10 years, given that he or she is from the West? (e) is from the West, given that he or she has been loyal less than 1 year? (f) is from the South, given that he or she has been loyal less than 1 year? (g) has been loyal 1 or more years, given that he or she is from the East? (h) has been loyal 1 or more years, given that he or she is from the West? (i) Are the events from the East and loyal 15 or more years independent? Explain.

USA Today gave the information shown in the table about ages of children receiving toys. The percentages represent all toys sold. What is the probability that a toy is purchased for someone (a) 6 years old or older? (b) 12 years old or younger? (c) between 6 and 12 years old? (d) between 3 and 9 years old? A child between 10 and 12 years old looks at this probability distribution and asks, "Why are people more likely to buy toys for kids older than I am (13 and over) than for kids in my age group (10-12)?" How would you respond? $$\begin{array}{l|c} \hline \text { Age (years) } & \text { Percentage of Toys } \\ \hline 2 \text { and under } & 15 \% \\ 3-5 & 22 \% \\ 6-9 & 270 / 6 \\ 10-12 & 14 \% \\ 13 \text { and over } & 22 \% \end{array}$$

Suppose two events \(A\) and \(B\) are mutually exclusive, with \(P(A) \neq 0\) and \(P(B) \neq 0 .\) By working through the following steps, you'll see why two mutually exclusive events are not independent. (a) For mutually exclusive events, can event \(A\) occur if event \(B\) has occurred? What is the value of \(P(A \mid B) ?\) (b) Using the information from part (a), can you conclude that events \(A\) and \(B\) are not independent if they are mutually exclusive? Explain.

There are 15 qualified applicants for 5 trainee positions in a fast-food management program. How many different groups of trainees can be selected?
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