/*! 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 8 Two cards are selected at random... [FREE SOLUTION] | 91Ó°ÊÓ

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Two cards are selected at random without replacement from a well-shuffled deck of 52 playing cards. Find the probability of the given event. Two cards of the same suit are drawn.

Short Answer

Expert verified
The probability of drawing two cards of the same suit from a well-shuffled deck of 52 playing cards, without replacement, is \(\frac{4}{17}\).

Step by step solution

01

Define the sample space and the favorable outcomes

We know the deck consists of 52 cards, and therefore there are a total of 52 first-card options. Once the first card is drawn, there are 51 cards remaining from which we can draw the second card. Since there are 4 suits and each suit has 13 cards, our favorable outcomes are drawing two cards from one suit.
02

Calculate the probability of drawing the first card from a specific suit (hearts, for example)

With 13 out of 52 cards belonging to hearts, the probability of drawing the first card from hearts would be \(\frac{13}{52}\), which simplifies to \(\frac{1}{4}\).
03

Calculate the probability of drawing the second card from the same suit (hearts)

After drawing one card from hearts, there are 12 hearts remaining and a total of 51 cards left in the deck. So the probability of drawing the second card from hearts, given that the first card is from hearts, is \(\frac{12}{51}\).
04

Calculate the probability of drawing two cards from the same suit (hearts)

Using the probabilities from Steps 2 and 3, the probability of drawing two cards from hearts is \(\frac{1}{4} \times \frac{12}{51} = \frac{1}{17}\).
05

Apply the same logic to the other three suits

Since the probabilities of drawing two cards from the same suit will be the same for diamonds, clubs, and spades, we can multiply the probability for hearts by 4. That is \(4 \times \frac{1}{17} = \frac{4}{17}\).
06

Write the final answer

The probability of drawing two cards of the same suit from a well-shuffled deck of 52 playing cards, without replacement, is \(\frac{4}{17}\).

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Key Concepts

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

Understanding Combinatorics in Card Games
Combinatorics is the branch of mathematics that deals with counting and arranging objects. In card games, it helps us calculate probabilities by understanding different combinations. When you draw cards from a deck, you're engaging with combinatorial concepts.

Drawing two cards from the same suit involves a specific combination. The number of ways to choose two cards out of the 52-card deck involves calculating combinations. For each suit (hearts, diamonds, clubs, spades), there are 13 cards. The goal is to determine how many combinations yield two cards of the same suit.
  • First Card: There are 13 choices for the first card of any specific suit.
  • Second Card: Once the first card is chosen, 12 cards remain in that suit.
Combinatorics transforms these choices into mathematical functions. It allows us to calculate the odds of complex outcomes in a logical and structured way. By understanding this, you can solve a broad range of probability problems in card games and beyond.
Calculating Card Probability
Card probability explores the likelihood of drawing certain cards from a deck. This probability is founded on the principles of combinatorics and is vital for strategic decision-making in card games.

When calculating the probability of drawing two cards of the same suit:
  • First Card Probability: Since there are 13 cards of one suit out of 52, the probability is \( \frac{13}{52} \) or \( \frac{1}{4} \).
  • Second Card Probability: With the first card drawn, only 12 of that suit are left among 51 cards in total, leading to a probability of \( \frac{12}{51} \).
By multiplying these probabilities, you find the chance of drawing two cards from the same suit. The complete chance of this happening for any of the four suits involves multiplying through all possible suits, i.e., \( 4 \times \frac{1}{17} = \frac{4}{17} \). Understanding this probability supports better decision-making in card strategies.
Enhancing Mathematics Education with Real-World Examples
Incorporating real-world examples, like card games, into mathematics education makes learning more engaging and effective. By using problems that involve common activities, students can connect abstract math concepts to their experiences.

Real-world scenarios, such as card probability, illustrate how mathematical theories apply to everyday life. This approach:
  • Builds Interest: Learners are often more motivated when they see the relevance of math in activities like playing cards.
  • Improves Understanding: Practical examples make abstract ideas clearer and more approachable.
  • Enhances Problem-Solving Skills: Students learn to apply logical reasoning to solve real challenges.
Educators can use card games and other daily examples to help students visualize and grasp mathematical concepts. Such techniques create more dynamic lessons and foster a deeper understanding of mathematics.

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Most popular questions from this chapter

Applicants for temporary office work at Carter Temporary Help Agency who have successfully completed a typing test are then placed in suitable positions by Nancy Dwyer and Darla Newberg. Employers who hire temporary help through the agency return a card indicating satisfaction or dissatisfaction with the work performance of those hired. From past experience it is known that \(80 \%\) of the employees placed by Nancy are rated as satisfactory, and \(70 \%\) of those placed by Darla are rated as satisfactory. Darla places \(55 \%\) of the temporary office help at the agency and Nancy the remaining \(45 \%\). If a Carter office worker is rated unsatisfactory, what is the probability that he or she was placed by Darla?

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