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Using five-element permutations as a sample space, determine the probability that a hand of five cards, chosen from an ordinary deck of 52 cards, will have all the cards from the same suit.

Short Answer

Expert verified
The probability of having all cards from the same suit is approximately 0.00198.

Step by step solution

01

Calculate Total Permutations of Five Cards

To find the number of ways to choose 5 cards from a 52-card deck, we look at the number of permutations. Choose any 5 cards from the 52 without caring about the order: \[\text{Total permutations} = 52 \times 51 \times 50 \times 49 \times 48\]
02

Calculate Favorable Permutations for Same Suit

A suit consists of 13 cards. We need to choose 5 cards from 13 cards of the same suit. Calculate permutations for one suit, then multiply by 4 (since there are 4 suits - hearts, diamonds, clubs, spades): \[\text{Permutations for one suit} = 13 \times 12 \times 11 \times 10 \times 9\]\[\text{Total permutations for all suits} = 4 \times (13 \times 12 \times 11 \times 10 \times 9)\]
03

Calculate Probability of All Cards from Same Suit

The probability is the ratio of favorable outcomes to the total number of outcomes:\[\text{Probability} = \frac{4 \times (13 \times 12 \times 11 \times 10 \times 9)}{52 \times 51 \times 50 \times 49 \times 48}\]
04

Simplify the Fraction for Probability

Simplify the fraction obtained in the previous step by calculating the actual numbers, or use a calculator to reach the final probability outcome.

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Key Concepts

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

Permutations
Permutations are a key concept in probability theory. They refer to the different ways in which a set or subset of items can be arranged. This concept is especially important when the order of items matters. For example, if you were interested in determining the different orders in which five books can be placed on a shelf, you would consider the permutations of these five books.

In the context of card probability, permutations help us understand the number of possible ways to arrange cards. For instance, if we have a deck of 52 cards and want to know how many ways we can arrange 5 cards out of this deck, permutations help us solve this problem. By calculating the total permutations, we get the number: \[52 \times 51 \times 50 \times 49 \times 48\]This represents the number of possible unique arrangements of five cards drawn without replacement.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, quality assessment, and arrangements within a set. It is crucial for calculating possibilities in various scenarios, which includes permutations and combinations. In permutations, unlike combinations, the order of elements is significant.

When card games are involved, combinatorics helps calculate the number of ways to choose a specific hand. By focusing on favorable permutations, like those of the same suit from a deck, we apply combinatorial techniques to determine possibilities. Take a hand of 5 cards from a 52-card deck, each suit having 13 cards: \[13 \times 12 \times 11 \times 10 \times 9\]This is repeated for each suit. To find total possibilities, combine permutations across all four suits, multiplying by 4 for the suits considered in the deck.
Card Probability
Card probability is a fascinating area of probability theory. It deals with the likelihood of drawing certain cards or combinations from a standard deck. A deck of cards is a great representation for learning probability because it consists of a finite number of elements with a well-defined structure.

To find the probability of drawing a hand of five cards all from the same suit, we use the concepts of permutations and combinatorics. This involves determining the favorable permutations over all possible permutations: \[\text{Probability} = \frac{4 \times (13 \times 12 \times 11 \times 10 \times 9)}{52 \times 51 \times 50 \times 49 \times 48}\]This formula gives us the chance, or probability, of drawing such a hand under standard conditions.
Sample Space
In probability theory, a sample space is the set of all possible outcomes of a random experiment. When dealing with cards, the sample space is the complete set of all potential hands or arrangements you can draw from the deck.

For a deck of 52 cards, every permutation of a five-card hand is a possible outcome in the sample space. By calculating this neatly defined sample space, like for a hand of five cards, we rely on permutations: \[52 \times 51 \times 50 \times 49 \times 48\]With this, the sample space provides the denominator in probability calculations — helping comprehend what portion of the whole set of possible outcomes produced a favorable hand, such as all cards of the same suit.

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Most popular questions from this chapter

Given an array \(A\) of length \(n\) (chosen from some set that has an underlying ordering), you can select the largest element of the array by first setting \(L=A[1]\) and then comparing \(L\) to the remaining elements of the array, one at a time, replacing \(L\) with \(A[i]\) if \(A[i]\) is larger than \(L\). Assume that the elements of \(A\) are randomly chosen. For \(i>1\), let \(X_{i}=1\) if an element \(i\) of \(A\) is larger than any element of \(A[1: i-1]\). Let \(X_{1}=1 .\) What does \(X_{1}+X_{2}+\cdots+X_{n}\) have to do with the number of times you assign a value to \(L ?\) What is the expected number of times you assign a value to \(L\) ?

How many five-card hands chosen from a standard deck of playing cards consist of five cards in a row (such as the nine of diamonds, ten of clubs, jack of clubs, queen of hearts, and king of spades)? Such a hand is called a straight. What is the probability that a five-card hand is a straight? Explore whether you get the same answer by using five- element sets as your model of hands or five-element permutations as your model of hands.

Prove as tight upper and lower bounds as you can for \(\sum_{i=1}^{k}(1 / i)\). For this purpose, it is useful to remember the definition of the natural logarithm as an integral involving \(1 / x\) and to draw rectangles and other geometric figures above and below the curve.

A candy machine in a school has d different kinds of candy. Assume (for simplicity) that all these kinds of candy are equally popular and there is a large supply of each. Suppose that c children come to the machine, and each child purchases one package of candy. One of the kinds of candy is a Snackers bar. a. What is the probability that any given child purchases a Snackers bar? b. Let Yi be the number of Snackers bars that Child i purchases—Yi is either 0 or 1. What is the expected value of Yi? c. Let Y be the random variable Y1 + Y2 +···+ Yc. What is the expected value of Y ? d. What is the expected number of Snackers bars that are purchased? e. Does the same result apply to any of the varieties of candy?

A die is made of a cube with a square painted on one side, a circle on two sides, and a triangle on three sides. If the die is rolled twice, what is the probability that the two shapes you see on top are the same?

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