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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 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? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?

Short Answer

Expert verified
The probabilities are as follows: a. The probability that a hand will consist entirely of spades is \(\frac{1287}{2598960}\). The probability that a hand will consist entirely of a single suit is \(4 * \frac{1287}{2598960}\). b. The probability that a hand consists entirely of spades and clubs with both suits represented is \(\frac{63206}{2598960}\). c. The probability that a hand contains cards from exactly two suits is \(6 * \frac{63206}{2598960}\).

Step by step solution

01

Probability of a hand consisting entirely of spades

The probability is calculated as the ratio of the number of favorable outcomes to the total number of outcomes. Here, it's known that there are 1,287 hands consisting entirely of spades out of a total of 2,598,960 hands. So, the probability is calculated as \(\frac{1287}{2598960}\)
02

Probability of a hand consisting entirely of a single suit

Remember that a deck of cards consists of 4 different suits: spades, hearts, diamonds, and clubs. So the probability of getting a hand consisting entirely of any single suit would be four times the probability of getting a hand consisting entirely of spades. So, the probability would be \(4 * \frac{1287}{2598960}\)
03

Probability of a hand consisting entirely of spades and clubs

Again, the probability is calculated as the ratio of the number of favorable outcomes to the total number of outcomes. Here, it's known that there are 63,206 hands consisting entirely of spades and clubs (with both suits represented) out of a total of 2,598,960 hands. So, the probability is calculated as \(\frac{63206}{2598960}\)
04

Probability of a hand containing cards from exactly two suits

A standard deck has 4 different suits, so there are \(\binom{4}{2} = 6\) ways to select 2 suits. Now from the results in Step 3, each has the probability \(\frac{63206}{2598960}\). So, the probability of a hand containing cards from exactly two suits would be \(6 * \frac{63206}{2598960}\)

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Key Concepts

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

Combinatorics
Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects. It helps determine the likelihood of different scenarios, especially in probability theory. In the context of card games, combinatorics is used to calculate the number of possible hands or outcomes.

When dealing cards, the order does not matter, so we use combinations rather than permutations. A combination considers the selection of objects without regard to order, which is essential in card games. The number of ways to choose a hand of cards can be calculated using the binomial coefficient, denoted as \( \binom{n}{k} \). Here, \(n\) is the total number of cards, and \(k\) is the number selected. This formula helps find how many ways you can deal a specific hand from a deck, underpinning probabilities in card games.
  • For example, the number of ways to pick 5 cards from a 52-card deck is given by \( \binom{52}{5} \).
This basic understanding of combinatorics forms the skeleton upon which probability calculations rest, as seen in card-based probability problems.
Cards Probability
Cards probability uses combinatorics to determine the chances of certain hands being dealt in a card game. The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes, both of which are figured using combinations.

For instance, if you want to know the probability of drawing a hand consisting entirely of spades, you first find the number of all-spade combinations using \( \binom{13}{5} \) because there are 13 spades in the deck of 52 cards. Then, this result is divided by the total number of 5 card combinations possible in a deck \( \binom{52}{5} \).
  • The formula for probability here is: \( \frac{\text{Number of favorable hands}}{\text{Total number of hands}} \).
By applying this logic, you can assess the likelihood of a hand consisting entirely of a single suit, or even combinations of two suits like spades and clubs.
Binomial Coefficient
The binomial coefficient, \( \binom{n}{k} \), is central to both combinatorics and probability. It indicates the number of ways to choose \(k\) elements from \(n\) elements without regard to order. Mathematically, it's defined as:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
  • "!" denotes factorial, which means the product of all positive integers up to that number.
  • This figure tells us how many possible combinations (hands) exist in a set of cards, enabling us to calculate probabilities accurately.
For card probabilities, understanding how binomial coefficients work is crucial. They allow us to map out every possible hand scenario, leading to precise probability estimates, whether for hands of a single suit or hands comprising exactly two suits. In exercises like these, applying the binomial coefficient is key to unlocking the broad range of potential outcomes in card games.

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Most popular questions from this chapter

A Gallup survey of 2002 adults found that \(46 \%\) of women and \(37 \%\) of men experience pain daily (San Luis Obispo Tribune, April 6, 2000). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the events selected adult is male and selected adult experiences pain daily independent or dependent? Explain.

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\) and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F ?\) b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F) ?\) A Venn diagram might help.)

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