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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

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 , 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 .\) " Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) b. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.7\) c. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) d. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\) e. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.4\) f. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\)

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a schedule is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is \(.8\). 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only . 6 . 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only .5. 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only \(.7 .\) Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

Of the 10,000 students at a certain university, 7000 have Visa cards, 6000 have MasterCards, and 5000 have both. Suppose that a student is randomly selected. a. What is the probability that the selected student has a Visa card? b. What is the probability that the selected student has both cards? c. Suppose you learn that the selected individual has a Visa card (was one of the 7000 with such a card). Now what is the probability that this student has both cards? d. Are the events has \(a\) Visa card and has a MasterCard independent? Explain. e. Answer the question posed in Part (d) if only 4200 of the students have both cards.

A single-elimination tournament with four players is to be held. In Game 1 , the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3 , the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: \(P(\) seed 1 defeats seed 4\()=.8\) \(P(\) seed 1 defeats seed 2\()=.6\) \(P(\) seed 1 defeats seed 3\()=.7\) \(P(\) seed 2 defeats seed 3\()=.6\) \(P(\) seed 2 defeats seed 4\()=.7\) \(P(\) seed 3 defeats seed 4\()=.6\) a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate Game 3 in the tournament? (This will depend on the outcomes of Games 1 and 2.) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is oyer 6 ft in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A) ?\) Why?

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