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A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are 2,598,960 possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 of the possible hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented?

Short Answer

Expert verified
The probability of a hand consisting entirely of spades is approximately 0.000495, while the probability of a hand consisting entirely of a single suit is approximately 0.00198. The probability of a hand consisting of only spades and clubs, with both suits represented, is approximately 0.02434.

Step by step solution

01

Find the total number of hands

To find the total number of 5-card hands that can be dealt from a deck of 52 cards, we will use the combination formula, which is given by: C(n, r) = n! / (r!(n-r)!), where n is the total number of items and r is the number of items we want to choose. In our case, n = 52 and r = 5, so: C(52, 5) = 52! / (5!(52 - 5)!) = \( \frac{52!}{5!47!} \)= 2,598,960 This tells us that there are 2,598,960 possible 5-card hands that can be dealt from a standard 52-card deck.
02

Calculate the probability of a hand consisting entirely of spades

We are given that there are only 1,287 hands consisting entirely of spades. To find the probability of getting a hand of all spades, we will divide the favorable outcome (1,287) by the total number of possible outcomes (2,598,960). P(hand of all spades) = \( \frac{1287}{2,598,960} \) ≈ 0.000495
03

Calculate the probability of a hand consisting entirely of a single suit

To calculate the probability of a hand consisting entirely of a single suit, we will multiply the probability of obtaining a hand of all spades by the number of suits (4). P(hand of a single suit) = 4 * P(hand of all spades) = 4 * 0.000495 ≈ 0.00198
04

Calculate the probability of a hand consisting of only spades and clubs, with both suits represented

We are given that there are exactly 63,206 possible hands containing only spades and clubs, with both suits represented. To find the probability of getting such a hand, we will divide the favorable outcome (63,206) by the total number of possible outcomes (2,598,960). P(hand of spades and clubs with both suits represented) = \( \frac{63,206}{2,598,960} \) ≈ 0.02434 Now we have the answers for both parts of the question: a) Probability of a hand consisting entirely of spades: ≈ 0.000495 Probability of a hand consisting entirely of a single suit: ≈ 0.00198 b) Probability of a hand consisting entirely of spades and clubs, with both suits represented: ≈ 0.02434

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Key Concepts

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

Combination Formula
Understanding the combination formula is crucial in the realm of probability and statistics. Simply put, the combination formula calculates the number of ways a certain number of items can be selected from a larger group without considering the order.

For example, if you're dealing with a deck of cards and want to know how many different 5-card hands you can draw, the combination formula is your go-to method. The formula you'll use is

\[ C(n, r) = \frac{n!}{r!(n-r)!} \]

where \( n \) stands for the total number of items to choose from (in a standard deck, 52 cards), and \( r \) is the number of items you're choosing (in this case, a 5-card hand). The exclamation mark \( ! \) represents the factorial, meaning you multiply the sequence of descending natural numbers from the number itself down to 1 (for instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)). This formula is essential in problems where the order does not matter, like selecting a random hand of cards.
Permutations and Combinations
Distinguishing between permutations and combinations is vital when tackling problems involving the arrangement of items. Permutations are all about arrangements where the order matters, while combinations focus on selections where the order is irrelevant.

The key difference is that permutations would consider the sequence of items as distinct, whereas combinations treat the same items in different sequences as the same selection. For example, if you pick two cards from a deck, getting an Ace then a King is treated differently than getting a King then an Ace in permutations, but not in combinations.

For our case with the deck of cards, because drawing the five of spades followed by the three of hearts is effectively the same hand as drawing the three of hearts followed by the five of spades, we use combinations (since the order in which the cards are drawn doesn't change the hand itself).
Favorable Outcomes
In probability, identifying the favorable outcomes—the results of an experiment that we're interested in—is a crucial step. These outcomes make up the numerator in the probability calculation. The more favorable outcomes you have, the higher the chances that the experiment will result in what you're expecting.

In the context of our card-dealing problem, the favorable outcomes for drawing a hand entirely of spades are the number of combinations that produce a 5-card hand from the 13 spades in the deck. If you calculate this number, it's 1,287. Think of it as sifting through all those thousands of possible 5-card hands and picking out the ones that give you exactly what you need: a handful of spades.

This concept isn't restricted to cards—it's a fundamental concept that applies to all kinds of probability questions, whether you're drawing cards, rolling dice, or picking marbles from a bag.
Total Outcomes
Total outcomes form the denominator in our probability fraction, representing the entire range of possible results from our random experiment. In the world of cards, we calculated earlier that there are a staggering 2,598,960 different 5-card combinations that can be drawn from a standard deck.

Whenever you tackle a probability problem, knowing or determining the total possible outcomes provides the context for how rare or common the favorable outcomes are. This is why the first step is often calculating the total number of possible combinations.

For any event, if there's only one way it can occur, and it does, your probability is 1—or a 100% certainty. But, as the number of total outcomes increases, the probability of any single event occurring typically goes down, assuming each outcome is equally likely. That's why it's less likely to draw a perfect hand of spades than to draw any given random hand.

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Most popular questions from this chapter

5.62 An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. In Exercise \(5.34,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. b. The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer.

The article "A Crash Course in Probability" from The Economist included the following information: The chance of being involved in an airplane crash when flying on an Airbus 330 from London to New York City on Virgin Atlantic Airlines is 1 in \(5,371,369 .\) This was interpreted as meaning that you "would expect to go down if you took this flight every day for 14,716 years." The article also states that a person could "expect to fly on the route for 14,716 years before plummeting into the Atlantic." Comment on why these statements are misleading.

A small college has 2700 students enrolled. Consider the chance experiment of selecting a student at random. For each of the following pairs of events, indicate whether or not you think they are mutually exclusive and explain your reasoning. a. the event that the selected student is a senior and the event that the selected student is majoring in computer science. b. the event that the selected student is female and the event that the selected student is majoring in computer science. c. the event that the selected student's college residence is more than 10 miles from campus and the event that the selected student lives in a college dormitory. d. the event that the selected student is female and the event that the selected student is on the college football team.

The National Center for Health Statistics (www.cdc .gov/nchs/data/nvsr/nvsr64/nvsr64_12.pdf, retrieved April 25,2017 ) gave the following information on births in the United States in 2014 : $$ \begin{array}{|lr|} \hline \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 3,848,214 \\ \text { Twins } & 135,336 \\ \text { Triplets } & 4,233 \\ \text { Quadruplets } & 246 \\ \text { Quintuplets or higher } & 47 \\ \hline \end{array} $$ Use this information to estimate the probability that a randomly selected pregnant woman who gave birth in 2014 a. delivered twins b. delivered quadruplets c. gave birth to more than a single child

In a particular state, automobiles that are more than 10 years old must pass a vehicle inspection in order to be registered. This state reports the probability that a car more than 10 years old will fail the vehicle inspection is 0.09 . Give a relative frequency interpretation of this probability.

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